Syllabus: Conetnt to study
a) Find the derivatives by definition of:- Inverse trigonometric Function
- Exponential Function
- logarithmic Functions
![Derivatives Exercise: 15.1 Class 12 Basic Mathematics Solution [NEB UPDATED]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgvlhO8qBWd9TZnN9YZNmzcIrOKszhUHv5bOUWlIV5PWacppRsxAExde6B3DtpHePhk105zm6xVbSE1hgmbo1lUINFtaLMlHF5TwkQpYCSG8223imgvVa20LoPadWPcgGqZ6CI23mXp3omsRVw7Ve1b2ZtPhosnYc-xgoJYv0ziav65CKH1Cgiju_a8/s1200/Exercise%20%2015.1.png "Derivatives Exercise: 15.1 Class 12 Basic Mathematics Solution [NEB UPDATED]")
Exercise 15.1
Find, From First principles, the derivatives of: (Ex 1-5)
1. (i) ${e^{\sqrt x }}$
f(x) = ${{\rm{e}}^{\sqrt {\rm{x}} }}$, f(x + h) = ${{\rm{e}}^{\sqrt {{\rm{x}}
+ {\rm{h}}} }}$.
f’(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{\sqrt {\rm{x}} }}}
\right)$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\rm{e}}^{\sqrt {{\rm{x}} + {\rm{h}}} }} - {{\rm{e}}^{\sqrt {\rm{x}}
}}}}{{\rm{h}}}$ ….(i)
Let $\sqrt {\rm{x}} $ = u and $\sqrt {{\rm{x}} + {\rm{h}}} $ = u + k.
So, that k = $\sqrt {{\rm{x}} + {\rm{h}}} $ - $\sqrt {\rm{x}} $.
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{\sqrt {\rm{x}} }}}
\right)$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{{\rm{e}}^{{\rm{u}} + {\rm{k}}}} - {{\rm{e}}^{\rm{u}}}}}{{\rm{h}}}$ =
h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{{\rm{e}}^{\rm{u}}}\left( {{{\rm{e}}^{\rm{k}}} - 1}
\right)}}{{\rm{h}}}$
= euh $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\left( {\frac{{{{\rm{e}}^{\rm{k}}} -
1}}{{\rm{k}}}.\frac{{\rm{k}}}{{\rm{h}}}} \right)$ = eu$\left(
{{\rm{k}} \to 0\frac{{{{\rm{e}}^{\rm{k}}} - 1}}{{\rm{k}}}} \right)\left(
{{\rm{h\: }}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\:
}}0\frac{{\rm{k}}}{{\rm{h}}}} \right)$
= eu.1 h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{\sqrt {{\rm{x}} + {\rm{h}}} - \sqrt {\rm{x}} }}{{\rm{h}}}$.
= ${{\rm{e}}^{\sqrt {\rm{x}} }}$h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sqrt
{{\rm{x}} + {\rm{h}}} - \sqrt {\rm{x}} }}{{\rm{h}}}{\rm{*}}\frac{{\sqrt
{{\rm{x}} + {\rm{h}}} + \sqrt {\rm{x}} }}{{\sqrt {{\rm{x}} +
{\rm{h}}} + \sqrt {\rm{x}} }}$${\rm{\: }}$= ${{\rm{e}}^{\sqrt {\rm{x}}
}}$h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{x}}
+ {\rm{h}} - {\rm{x}}}}{{{\rm{h}}\left( {\sqrt {{\rm{x}} + {\rm{h}}} +
\sqrt {\rm{x}} } \right)}}$ = ${{\rm{e}}^{\sqrt {\rm{x}} }}$h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\rm{h}}}{{{\rm{h}}\left( {\sqrt {{\rm{x}} + {\rm{h}}} + \sqrt
{\rm{x}} } \right)}}$
= ${{\rm{e}}^{\sqrt {\rm{x}} }}$h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{1}{{\sqrt
{{\rm{x}} + {\rm{h}}} + \sqrt {\rm{x}} }}$ = ${{\rm{e}}^{\sqrt {\rm{x}}
}}$$\frac{1}{{\sqrt {\rm{x}} + \sqrt {\rm{x}} }}$ =
$\frac{{{{\rm{e}}^{\sqrt {\rm{x}} }}}}{{2\sqrt {\rm{x}} }}$.
(ii) ${e^{\sin x}}$
f(x) = ${{\rm{e}}^{{\rm{sinx}}}}$, f(x + h) = ${{\rm{e}}^{{\rm{sin}}\left(
{{\rm{x}} + {\rm{h}}} \right)}}$.
f’(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{{\rm{sinx}}}}} \right)$ =
x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\rm{e}}^{{\rm{sin}}\left( {{\rm{x}} + {\rm{h}}} \right)}} -
{{\rm{e}}^{{\rm{sinx}}}}}}{{\rm{h}}}$ ….(i)
Let sinx = u and sin(x + h) = u + k.
So, that k = sin(x + h) – sinx.
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{{\rm{sinx}}}}} \right)$ =
h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{{\rm{e}}^{{\rm{u}} + {\rm{k}}}} - {{\rm{e}}^{\rm{u}}}}}{{\rm{h}}}$ =
h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{{\rm{e}}^{\rm{u}}}\left( {{{\rm{e}}^{\rm{k}}} - 1}
\right)}}{{\rm{h}}}$
= euh $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\left( {\frac{{{{\rm{e}}^{\rm{k}}} - 1}}{{\rm{k}}}.\frac{{\rm{k}}}{{\rm{h}}}}
\right)$ = eu$\left( {{\rm{k}} \to 0\frac{{{{\rm{e}}^{\rm{k}}} -
1}}{{\rm{k}}}} \right)\left( {{\rm{h\:
}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\:
}}0\frac{{\rm{k}}}{{\rm{h}}}} \right)$
= eu.1 h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{\sin \left( {{\rm{x}} + {\rm{h}}} \right) -
{\rm{sinx}}}}{{\rm{h}}}$.
= ${{\rm{e}}^{{\rm{sinx}}}}$h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2\cos
\frac{{{\rm{x}} + {\rm{h}} + {\rm{x}}}}{2}.\sin \frac{{{\rm{x}} + {\rm{h}} -
{\rm{x}}}}{2}}}{{\rm{h}}}$.
${\rm{\: }}$= ${{\rm{e}}^{{\rm{sinx}}}}$h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2\cos \left(
{{\rm{x}} + \frac{{\rm{h}}}{2}} \right)\sin
\frac{{\rm{h}}}{2}}}{{\rm{h}}}$
= ${{\rm{e}}^{{\rm{sinx}}}}$h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $2\cos \left(
{{\rm{x}} + \frac{{\rm{h}}}{2}} \right)$.$\frac{{\sin
\frac{{\rm{h}}}{2}}}{{\frac{{\rm{h}}}{2}}}.\frac{1}{2}$
= ${{\rm{e}}^{{\rm{sinx}}}}$cosx.1 =
${{\rm{e}}^{{\rm{sinx}}}}$.cosx.
(iii) ${e^{\tan x}}$
f(x) = ${{\rm{e}}^{{\rm{tanx}}}}$, f(x + h) = ${{\rm{e}}^{{\rm{tan}}\left(
{{\rm{x}} + {\rm{h}}} \right)}}$.
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{{\rm{tanx}}}}} \right)$ =
x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\rm{e}}^{{\rm{tan}}\left( {{\rm{x}} + {\rm{h}}} \right)}} -
{{\rm{e}}^{{\rm{tanx}}}}}}{{\rm{h}}}$ ….(i)
Let tanx = u and tan(x + h) = u + k.
So, that k = tan(x + h) – tanx.
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{{\rm{tanx}}}}} \right)$ =
h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{{\rm{e}}^{{\rm{u}} + {\rm{k}}}} - {{\rm{e}}^{\rm{u}}}}}{{\rm{h}}}$ =
h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{{\rm{e}}^{\rm{u}}}\left( {{{\rm{e}}^{\rm{k}}} - 1}
\right)}}{{\rm{h}}}$
= euh $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left(
{\frac{{{{\rm{e}}^{\rm{k}}} - 1}}{{\rm{k}}}.\frac{{\rm{k}}}{{\rm{h}}}}
\right)$
= eu$\left( {{\rm{k}} \to 0\frac{{{{\rm{e}}^{\rm{k}}} - 1}}{{\rm{k}}}}
\right)\left( {{\rm{h\:
}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\:
}}0\frac{{\rm{k}}}{{\rm{h}}}} \right)$
= etanx.1 h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{\tan \left( {{\rm{x}} + {\rm{h}}} \right) -
{\rm{tanx}}}}{{\rm{h}}}$.
= ${{\rm{e}}^{{\rm{tanx}}}}$h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\frac{{\sin
\left( {{\rm{x}} + {\rm{h}}} \right)}}{{\cos \left( {{\rm{x}} + {\rm{h}}}
\right)}} - \frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}}}{{\rm{h}}}$.
= ${{\rm{e}}^{{\rm{tanx}}}}$h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sin \left(
{{\rm{x}} + {\rm{h}}} \right).{\rm{cosx}} - \cos \left( {{\rm{x}} + {\rm{h}}}
\right).{\rm{sinx}}}}{{{\rm{h}}.{\rm{cosx}}.{\rm{cos}}\left( {{\rm{x}} +
{\rm{h}}} \right)}}$${\rm{\: }}$= ${{\rm{e}}^{{\rm{tanx}}}}$h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{sin}}\left( {{\rm{x}} + {\rm{h}} - {\rm{x}}}
\right)}}{{{\rm{cosx}}.{\rm{cos}}\left( {{\rm{x}} + {\rm{h}}} \right)}}$
= ${{\rm{e}}^{{\rm{tanx}}}}$h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{1}{{{\rm{cosx}}.\cos \left( {{\rm{x}} + {\rm{h}}}
\right)}}$.$\frac{{\sin {\rm{h}}}}{{\rm{h}}}$.
=
${{\rm{e}}^{{\rm{tanx}}}}$.1.$\frac{1}{{{\rm{cosx}}.{\rm{cosx}}}}$.
= sec2x.etanx.
2. (i) $\log (\sin {\rm{ }}\frac{x}{a})$
Solution:
f(x) = log $\left( {\sin \frac{{\rm{x}}}{{\rm{a}}}} \right)$, f(x + h) =
log.sin $\frac{{{\rm{x}} + {\rm{h}}}}{{\rm{a}}}$.
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\log \left( {\sin
\frac{{\rm{x}}}{{\rm{a}}}} \right)} \right)$ = x
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\log
.{\rm{sin\: }}\frac{{{\rm{x}} + {\rm{h}}}}{{\rm{a}}} - {\rm{log}}\sin
\frac{{\rm{x}}}{{\rm{a}}}}}{{\rm{h}}}$ ….(i)
Let sin $\frac{{\rm{x}}}{{\rm{a}}}$ = u and sin $\frac{{{\rm{x}} +
{\rm{h}}}}{{\rm{a}}}$ = u + k, so that,
So, that k = sin $\frac{{{\rm{x}} + {\rm{h}}}}{{\rm{a}}}$ – sin
$\frac{{\rm{x}}}{{\rm{a}}}$.
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\log \left( {\sin
\frac{{\rm{x}}}{{\rm{a}}}} \right)} \right)$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{log}}.\left( {{\rm{u}} + {\rm{k}}} \right) -
{\rm{logu}}}}{{\rm{h}}}$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0$\frac{{\log \left(
{\frac{{{\rm{u}} + {\rm{k}}}}{{\rm{u}}}} \right)}}{{\rm{h}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\log
\left( {1 + \frac{{\rm{k}}}{{\rm{u}}}}
\right)}}{{\frac{{\rm{k}}}{{\rm{u}}}}}.\frac{{\left(
{\frac{{\rm{k}}}{{\rm{u}}}} \right)}}{{\rm{h}}}$ = k
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\log \left( {1
+ \frac{{\rm{k}}}{{\rm{u}}}} \right)}}{{\frac{{\rm{k}}}{{\rm{u}}}}}$. h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\rm{k}}}{{{\rm{h}}.{\rm{u}}}}$.
= $\frac{1}{{\rm{u}}}$.1. h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0$\frac{{\sin
\frac{{{\rm{x}} + {\rm{h}}}}{{\rm{a}}} - \sin
\frac{{\rm{x}}}{{\rm{a}}}}}{{\rm{h}}}$.
= $\frac{1}{{\rm{u}}}$h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{2\cos \frac{{2{\rm{x}} + {\rm{h}}}}{{2{\rm{a}}}}.\sin
\frac{{\rm{h}}}{{2{\rm{a}}}}}}{{\rm{h}}}$.
= $\frac{1}{{\rm{u}}}$h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 2.cos $\frac{{2{\rm{x}} + {\rm{h}}}}{{2{\rm{a}}}}$. $\frac{{\sin
\frac{{\rm{h}}}{{2{\rm{a}}}}}}{{\frac{{\rm{h}}}{{2{\rm{a}}}}.2{\rm{a}}}}$ =
$\frac{1}{{\rm{u}}}$. 2cos $\frac{{\rm{x}}}{{\rm{a}}}$. 1.
$\frac{1}{{2{\rm{a}}}}$.
= $\frac{1}{{\rm{a}}}$. $\frac{1}{{\sin \frac{{\rm{x}}}{{\rm{a}}}}}$.cos
$\frac{{\rm{x}}}{{\rm{a}}}$ = $\frac{1}{{\rm{a}}}$.cot
$\frac{{\rm{x}}}{{\rm{a}}}$.
(ii) $\log (\tan x)$
Solution:
f(x) = log.tanx, f(x + h) = log.tan (x + h).
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{log}}.{\rm{tanx}}} \right)$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\log \tan
\left( {{\rm{x}} + {\rm{h}}} \right) - \log {\rm{tanx}}}}{{\rm{h}}}$ ….(i)
Let tanx = u and tan (x + h) = u + k, so that,
So, that k = tan(x + h) – tanx.
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{log}}.{\rm{tanx}}} \right)$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{log}}.\left( {{\rm{u}} + {\rm{k}}} \right) -
{\rm{logu}}}}{{\rm{h}}}$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{1}{{\rm{h}}}$.$\log \left( {\frac{{{\rm{u}} + {\rm{k}}}}{{\rm{u}}}}
\right)$
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{1}{{\rm{h}}}$.log $\left( {1 + \frac{{\rm{k}}}{{\rm{u}}}} \right)$ = k
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\log \left( {1
+ \frac{{\rm{k}}}{{\rm{u}}}} \right)}}{{\frac{{\rm{k}}}{{\rm{u}}}}}$. h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\rm{k}}}{{{\rm{h}}.{\rm{u}}}}$.
= $\frac{1}{{\rm{u}}}$.1. h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0$\frac{{\tan \left(
{{\rm{x}} + {\rm{h}}} \right) - {\rm{tanx}}}}{{\rm{h}}}$.
= $\frac{1}{{\rm{u}}}$h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{\frac{{\sin \left( {{\rm{x}} + {\rm{h}}} \right)}}{{\cos \left(
{{\rm{x}} + {\rm{h}}} \right)}} -
\frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}}}{{\rm{h}}}$.
= $\frac{1}{{\rm{u}}}$h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{\sin \left( {{\rm{x}} + {\rm{h}}} \right).{\rm{cosx}} - \cos \left(
{{\rm{x}} + {\rm{h}}} \right).{\rm{sinx}}}}{{{\rm{hcos}}\left( {{\rm{x}} +
{\rm{h}}} \right).{\rm{cosx}}}}$
= $\frac{1}{{\rm{u}}}$. h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sin
\left( {{\rm{x}} + {\rm{h}} - {\rm{x}}} \right)}}{{{\rm{h}}.\cos \left(
{{\rm{x}} + {\rm{h}}} \right).{\rm{cosx}}}}$.
= $\frac{1}{{{\rm{tanx}}}}$ h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{sinh}}}}{{\rm{h}}}.\frac{1}{{\cos \left( {{\rm{x}} + {\rm{h}}}
\right)}}.\frac{1}{{{\rm{cosx}}}}$
= $\frac{1}{{{\rm{tanx}}}}.1.{\rm{\: \: }}\frac{1}{{{\rm{cosx}}}}.{\rm{\:
}}\frac{1}{{{\rm{cosx}}}}$
= $\frac{2}{{{\rm{sin}}2{\rm{x}}}}$ = 2cosec2x.
(iii) $\log (\sec {x^2})$
Solution:
f(x) = log.secx2, f(x + h) = log.sec(x + h)2.
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{log}}.{\rm{sec}}{{\rm{x}}^2}}
\right)$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\log \sec {{\left( {{\rm{x}} + {\rm{h}}} \right)}^2} - \log
{\rm{sec}}{{\rm{x}}^2}}}{{\rm{h}}}$ ….(i)
Let secx2 = u and sec (x + h)2 = u + k, so
that,
So, that k = sec (x + h)2 – secx2.
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{log}}.{\rm{sec}}{{\rm{x}}^2}}
\right)$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{log}}.\left( {{\rm{u}} + {\rm{k}}} \right) -
{\rm{logu}}}}{{\rm{h}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\log
\left( {1 + \frac{{\rm{k}}}{{\rm{u}}}}
\right)}}{{\frac{{\rm{k}}}{{\rm{u}}}}}$.$\frac{{{\rm{ku}}}}{{\rm{h}}}$.
=k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\log \left(
{1 + \frac{{\rm{k}}}{{\rm{u}}}} \right)}}{{\frac{{\rm{k}}}{{\rm{u}}}}}$.
h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\rm{k}}}{{{\rm{h}}.{\rm{u}}}}$.
=1.$\frac{1}{{\rm{u}}}$h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{\sec {{\left( {{\rm{x}} + {\rm{h}}} \right)}^2} -
{\rm{sec}}{{\rm{x}}^2}}}{{\rm{h}}}$.
= $\frac{1}{{\rm{u}}}$. h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\frac{1}{{\cos {{\left( {{\rm{x}} + {\rm{h}}} \right)}^2}}} -
\frac{1}{{\cos {{\rm{x}}^2}}}}}{{\rm{h}}}$.
= $\frac{1}{{\rm{u}}}$h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{{\rm{cos}}{{\rm{x}}^2} - \cos {{\left( {{\rm{x}} + {\rm{h}}}
\right)}^2}}}{{{\rm{hcos}}{{\left( {{\rm{x}} + {\rm{h}}}
\right)}^2}.{\rm{cos}}{{\rm{x}}^2}}}$
= $\frac{1}{{\rm{u}}}$h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{2\sin \frac{{{{\rm{x}}^2} + {{\left( {{\rm{x}} + {\rm{h}}}
\right)}^2}}}{2}.\sin \frac{{{{\left( {{\rm{x}} + {\rm{h}}} \right)}^2} -
{{\rm{x}}^2}}}{2}}}{{{\rm{h}}.{\rm{cos}}{{\rm{x}}^2}.\cos {{\left( {{\rm{x}} +
{\rm{h}}} \right)}^2}}}$
= $\frac{1}{{\rm{u}}}$h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{2\sin \left( {\frac{{2{{\rm{x}}^2} + 2{\rm{xh}} + {{\rm{h}}^2}}}{2}}
\right).\sin \left( {\frac{{2{\rm{xh}} + {{\rm{h}}^2}}}{2}}
\right)}}{{{\rm{h}}.{\rm{cos}}{{\rm{x}}^2}.\cos {{\left( {{\rm{x}} + {\rm{h}}}
\right)}^2}}}$
= $\frac{1}{{\rm{u}}}$ h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{2\sin \left( {{{\rm{x}}^2} + {\rm{xh}} +
\frac{{{{\rm{h}}^2}}}{2}} \right).\frac{{\sin \left( {{\rm{x}} +
\frac{{\rm{h}}}{2}} \right)}}{{{\rm{h}}\left( {{\rm{x}} + \frac{{\rm{h}}}{2}}
\right)}}.{\rm{h}}\left( {{\rm{x}} + \frac{{\rm{h}}}{2}}
\right)}}{{{\rm{h}}.{\rm{cos}}{{\rm{x}}^2}.\cos {{\left( {{\rm{x}} + {\rm{h}}}
\right)}^2}}}$
= $\frac{1}{{\rm{u}}}.\frac{{2{{\sin }^2}{\rm{x}}.1.{\rm{x}}}}{{\cos
{{\rm{x}}^2}.{\rm{cos}}{{\rm{x}}^2}}}$ =
$\frac{1}{{{\rm{sec}}{{\rm{x}}^2}}}.\frac{{2{\rm{x}}.{\rm{sin}}{{\rm{x}}^2}}}{{{\rm{cos}}{{\rm{x}}^2}.{\rm{cos}}{{\rm{x}}^2}}}$
= 2xtanx2.
3. (i) ${{\mathop{\rm Cos}\nolimits} ^{ - 1}}x$
Solution:
f(x) = cos-1x, f(x + h) = cos-1(x + h).
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\cos }^{ - 1}}{\rm{x}}}
\right)$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\cos }^{ - 1}}\left( {{\rm{x}} + {\rm{h}}} \right) - {{\cos }^{ -
1}}{\rm{x}}}}{{\rm{h}}}$ ….(i)
Let cos-1x = u and cos-1 (x + h) = u + k, so
that,
So, that x = cos u and x + h = cos(u + k).
h = cos(u + k) – cosu.
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\cos }^{ - 1}}{\rm{x}}} \right)$ =
k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0$\frac{{{\rm{u}} +
{\rm{k}} - {\rm{u}}}}{{\cos \left( {{\rm{u}} + {\rm{k}}} \right) -
{\rm{cosu}}}}$
= k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\rm{k}}}{{
- 2\sin \frac{{{\rm{u}} + {\rm{k}} + {\rm{u}}}}{2}.\sin
\frac{{\rm{k}}}{2}}}$
= k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\frac{{\rm{k}}}{2}}}{{ - \sin \left( {{\rm{u}} + \frac{{\rm{k}}}{2}}
\right).\sin \frac{{\rm{k}}}{2}}}$ = $ - \frac{1}{{{\rm{sinu}}}}$ = $ -
\frac{1}{{\sqrt {1 - {{\cos }^2}{\rm{u}}} }}$ = $ - \frac{1}{{\sqrt {1 -
{{\rm{x}}^2}} }}$.
(ii)${\tan ^{ - 1}}x$
Solution:
f(x) = tan-1x, f(x + h) = tan-1(x + h).
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\tan }^{ - 1}}{\rm{x}}} \right)$ =
h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{{\tan }^{ -
1}}\left( {{\rm{x}} + {\rm{h}}} \right) - {{\tan }^{ -
1}}{\rm{x}}}}{{\rm{h}}}$ ….(i)
Lettan-1x = u and tan-1 (x + h) = u + k, so
that,
So, that x = tan u and x + h = tan(u + k).
h = tan(u + k) – tanu.
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\tan }^{ - 1}}{\rm{x}}} \right)$ =
k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0$\frac{{{\rm{u}} +
{\rm{k}} - {\rm{u}}}}{{\tan \left( {{\rm{u}} + {\rm{k}}} \right) -
{\rm{tan}}.{\rm{u}}}}$
= k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\rm{k}}}{{\frac{{\sin \left( {{\rm{u}} + {\rm{k}}} \right)}}{{\cos
\left( {{\rm{u}} + {\rm{k}}} \right)}} -
\frac{{{\rm{sinu}}}}{{{\rm{cosu}}}}}}$
= k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{k}}.{\rm{cosu}}.{\rm{cos}}\left( {{\rm{u}} + {\rm{k}}}
\right)}}{{\sin \left( {{\rm{u}} + {\rm{k}}} \right).{\rm{cosu}} - \cos \left(
{{\rm{u}} + {\rm{k}}} \right).{\rm{sinu}}}}$
= k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{k}}.{\rm{cosu}}.{\rm{cos}}\left( {{\rm{u}} + {\rm{k}}}
\right)}}{{{\rm{sin}}\left( {{\rm{u}} + {\rm{k}} - {\rm{u}}} \right)}}$
= k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{k}}.{\rm{cosu}}.{\rm{cos}}\left( {{\rm{u}} + {\rm{k}}}
\right)}}{{{\rm{sink}}}}$
= cosu.cosu = cos2u
= $\frac{1}{{{{\sec }^2}{\rm{u}}}} = \frac{1}{{1 + {{\tan }^2}{\rm{u}}}} =
\frac{1}{{1 + {{\rm{x}}^2}}}$.
(iii) $\log ({{\mathop{\rm Cos}\nolimits} ^{ - 1}}x)$
Solution:
f(x) = log(cos-1x), f(x + h) = log(cos-1(x + h)).
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\log .{{\cos }^{ - 1}}{\rm{x}}}
\right)$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\log .{{\cos }^{ - 1}}\left( {{\rm{x}} + {\rm{h}}} \right) - \log
.{{\cos }^{ - 1}}{\rm{x}}}}{{\rm{h}}}$ ….(i)
Let cos-1x = u and cos-1 (x + h) = u + k, so
that,
So, that x = cos u and x + h = cos(u + k).
h = cos(u + k) – cosu.
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{log}}.{{\cos }^{ - 1}}{\rm{x}}}
\right)$ = k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{\log \left( {{\rm{u}} + {\rm{k}}} \right) -
{\rm{logu}}}}{{\rm{h}}}$
= k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\log
\left( {1 + \frac{{\rm{k}}}{{\rm{u}}}}
\right)}}{{\frac{{\rm{k}}}{{\rm{u}}}}}$. k
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\frac{{\rm{k}}}{{\rm{u}}}}}{{\rm{h}}}$.
= 1.k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\rm{k}}}{{\cos \left( {{\rm{u}} + {\rm{k}}} \right) -
{\rm{cosu}}}}$
= $\frac{1}{{\rm{u}}}$. k
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\rm{k}}}{{2\sin \frac{{{\rm{u}} + {\rm{k}} + {\rm{u}}}}{2}.\sin
\frac{{{\rm{u}} - {\rm{u}} - {\rm{k}}}}{2}}}$
= $\frac{1}{{\rm{u}}}$. k
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\rm{k}}}{{ -
2\sin \left( {{\rm{u}} + \frac{{\rm{k}}}{2}} \right).\sin
\frac{{\rm{k}}}{2}}}$
= $\frac{1}{{\rm{u}}}$. k
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\frac{{\rm{k}}}{2}}}{{\sin \left( {{\rm{u}} + \frac{{\rm{k}}}{2}}
\right).\sin \frac{{\rm{k}}}{2}}}$ = $ -
\frac{1}{{\rm{u}}}.\frac{1}{{{\rm{sinu}}}}$
= $ - \frac{1}{{{{\cos }^{ - 1}}{\rm{x}}}}.\frac{1}{{\sqrt {1 - {{\cos
}^2}{\rm{u}}} }}$ = $ - \frac{1}{{{{\cos }^{ - 1}}{\rm{x}}}}.\frac{1}{{\sqrt
{1 - {{\rm{x}}^2}} }}$.
= $ - \frac{1}{{{{\cos }^{ - 1}}{\rm{x}}\sqrt {1 - {{\rm{x}}^2}} }}$.
4. (i) ${x^x}$
Solution:
f(x) = xx = ${{\rm{e}}^{\log {{\rm{x}}^{\rm{x}}}}}$ =
ex.logx, f(x + h) = ${{\rm{e}}^{\left( {{\rm{x}} + {\rm{h}}}
\right)\left( {\log \left( {{\rm{x}} + {\rm{h}}} \right)} \right)}}$
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^{\rm{x}}}} \right)$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\rm{e}}^{\left( {{\rm{x}} + {\rm{h}}} \right).\log \left( {{\rm{x}}
+ {\rm{h}}} \right)}} - {{\rm{e}}^{{\rm{x}}.{\rm{logx}}}}}}{{\rm{h}}}$
….(i)
Let x.logx = u and (x + h).log(x + h) = u + k, so that,
So, that k = (x + h).log(x + h) – x.logx
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^{\rm{x}}}} \right)$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{{\rm{e}}^{{\rm{u}} + {\rm{k}}}} -
{{\rm{e}}^{\rm{u}}}}}{{\rm{h}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\rm{e}}^{\rm{u}}}\left( {{{\rm{e}}^{\rm{k}}} - 1}
\right)}}{{\rm{h}}}$.
= eu. h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\rm{e}}^{\rm{k}}} - 1}}{{\rm{k}}}.\frac{{\rm{k}}}{{\rm{h}}}$
=eu$\left({{\rm{k\: }}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}0{\rm{\: }}\frac{{{{\rm{e}}^{\rm{k}}} - 1}}{{\rm{k}}}} \right)\left( {{\rm{h\: }}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\: }}0\frac{{\rm{k}}}{{\rm{h}}}} \right)$
= eu.1. h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\left( {{\rm{x}} + {\rm{h}}} \right)\log \left( {{\rm{x}} + {\rm{h}}}
\right) - {\rm{x}}.{\rm{logx}}}}{{\rm{h}}}$
= elogx. h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{x}}.\log \left( {{\rm{x}} + {\rm{h}}} \right) + {\rm{hlog}}\left(
{{\rm{x}} + {\rm{h}}} \right) - {\rm{xlogx}}}}{{\rm{h}}}$
= xxh $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{x}}\left\{ {\log \left( {{\rm{x}} + {\rm{h}}} \right) -
{\rm{logx}}} \right\} + {\rm{h}}.\log \left( {{\rm{x}} + {\rm{h}}}
\right)}}{{\rm{h}}}$
= xxh $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\left\{ {\frac{{\rm{x}}}{{\rm{h}}}.\log \left( {\frac{{{\rm{x}} +
{\rm{h}}}}{{\rm{x}}}} \right) + \log \left( {{\rm{x}} + {\rm{h}}} \right)}
\right\}$.
= xxh $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\left\{ {\frac{{\log \left( {1 + \frac{{\rm{h}}}{{\rm{x}}}}
\right)}}{{\frac{{\rm{h}}}{{\rm{x}}}}} + \log \left( {{\rm{x}} + {\rm{h}}}
\right)} \right\}$ = xx (1 + logx)
(ii) ${2^{{x^2}}}$
Solution:
f(x) = ${2^{\rm{x}}}^2$ = ${{\rm{e}}^{\log {2^{\rm{x}}}^2}}$ =
${{\rm{e}}^{{{\rm{x}}^2}{\rm{log}}2}}$ ,f(x + h) = ${{\rm{e}}^{{{\left(
{{\rm{x}} + {\rm{h}}} \right)}^2}{\rm{log}}2}}$
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^{\rm{x}}}} \right)$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\rm{e}}^{{{\left( {{\rm{x}} + {\rm{h}}} \right)}^2}.\log 2}} -
{{\rm{e}}^{{{\rm{x}}^2}.{\rm{log}}2}}}}{{\rm{h}}}$ ….(i)
Let x2.log2 = u and (x + h)2.log2 = u + k, so that,
So, that k = (x + h)2.log2 – x2.log2.
= log2{x2 + 2xh + h2 – x2} =
log2h(2x + h)
When h → 0, k → 0. Then (i) becomes
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{2^{\rm{x}}}^2} \right)$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{{\rm{e}}^{{\rm{u}} + {\rm{k}}}} -
{{\rm{e}}^{\rm{u}}}}}{{\rm{h}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\rm{e}}^{\rm{u}}}\left( {{{\rm{e}}^{\rm{k}}} - 1}
\right)}}{{\rm{h}}}$.
= eu. h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\rm{e}}^{\rm{k}}} - 1}}{{\rm{k}}}.\frac{{\rm{k}}}{{\rm{h}}}$
= eu$\left( {{\rm{h\:
}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}0{\rm{\:
}}\frac{{{{\rm{e}}^{\rm{k}}} - 1}}{{\rm{k}}}} \right)\left( {{\rm{h\:
}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\:
}}0\frac{{\rm{k}}}{{\rm{h}}}} \right)$
= eu.1. h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{log}}2{\rm{h}}\left( {2{\rm{x}} + {\rm{h}}} \right)}}{{\rm{h}}}$
= eu.log2.2x
= ${{\rm{e}}^{{{\rm{x}}^2}.{\rm{log}}2}}$.log.2.2x =
2xlog2.${2^{\rm{x}}}^2$.${\rm{\: }}$
(iii) $\log ({x^x})$
Solution:
f(x) = log xx = x.logx , f(x + h) = (x + h)log(x + h).
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{log\:
}}{{\rm{x}}^{\rm{x}}}{\rm{\: }}} \right)$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left(
{{\rm{x}} + {\rm{h}}} \right)\log \left( {{\rm{x}} + {\rm{h}}} \right) -
{\rm{xlogx}}}}{{\rm{h}}}$ ….(i)
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{x}}.\log \left( {{\rm{x}} + {\rm{h}}} \right) + {\rm{h}}.\log
\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{x}}.{\rm{logx}}}}{{\rm{h}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{x}}\left\{ {\log \left( {{\rm{x}} + {\rm{h}}} \right) -
{\rm{logx}}} \right\} + {\rm{h}}.\log \left( {{\rm{x}} + {\rm{h}}}
\right)}}{{\rm{h}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left\{
{\frac{{\rm{x}}}{{\rm{h}}}\log \left( {\frac{{{\rm{x}} + {\rm{h}}}}{{\rm{x}}}}
\right) + \log \left( {{\rm{x}} + {\rm{h}}} \right)} \right\}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left\{
{\frac{{\log \left( {1 + \frac{{\rm{h}}}{{\rm{x}}}}
\right)}}{{\frac{{\rm{h}}}{{\rm{x}}}}} + \log \left( {{\rm{x}} + {\rm{h}}}
\right)} \right\}$ = 1 + logx.
5. (i) Sin x2
Solution:
f(x) = sin x2 = x.logx , f(x + h) = sin(x +
h)2.
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{sin\: }}{{\rm{x}}^2}} \right)$ =
h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sin
{{\left( {{\rm{x}} + {\rm{h}}} \right)}^2} -
{\rm{sin}}{{\rm{x}}^2}}}{{\rm{h}}}$ ….(i)
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2\cos
\frac{{{{\left( {{\rm{x}} + {\rm{h}}} \right)}^2} + {{\rm{x}}^2}}}{2}.\sin
\frac{{{{\left( {{\rm{x}} + {\rm{h}}} \right)}^2} -
{{\rm{x}}^2}}}{2}}}{{\rm{h}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2\cos
\frac{{2{{\rm{x}}^2} + 2{\rm{xh}} + {{\rm{h}}^2}}}{2}.\sin \frac{{2{\rm{xh}} +
{{\rm{h}}^2}}}{2}}}{{\rm{h}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2\cos
\left( {{{\rm{x}}^2} + {\rm{xh}} + \frac{{{{\rm{h}}^2}}}{2}} \right).\sin
{\rm{h}}\left( {{\rm{x}} + \frac{{\rm{h}}}{2}} \right)}}{{\rm{h}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 2cos $\left(
{{{\rm{x}}^2} + {\rm{xh}} + \frac{{{{\rm{h}}^2}}}{2}} \right)$. $\frac{{\sin
\left( {{\rm{x}} + \frac{{\rm{h}}}{2}} \right)}}{{{\rm{h}}\left( {{\rm{x}} +
\frac{{\rm{h}}}{2}} \right)}}$(x + $\frac{{\rm{h}}}{2}$)
= 2.cosx2.1.x = 2x.cosx2.
(ii) Sin (log x)
Solution:
f(x) = sin(logx) = x.logx , f(x + h) = sin(log(x + h))
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\sin \left( {{\rm{logx}}} \right)}
\right)$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\sin \left( {\log \left( {{\rm{x}} + {\rm{h}}} \right)}
\right)---{\rm{sin}}\left( {{\rm{logx}}} \right)}}{{\rm{h}}}$ ….(i)
Put logx = u and log(x + h) = u + k , so that,
k = log(x + h) – logx
When h → 0, k → 0, Then (i) becomes.
Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\sin \left( {{\rm{logx}}} \right)}
\right)$= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\sin \left( {{\rm{u}} + {\rm{k}}} \right) -
{\rm{sinu}}}}{{\rm{h}}}$.
= $\left( {{\rm{k\:
}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\: }}0\frac{{\sin
\left( {{\rm{u}} + {\rm{k}}} \right) - {\rm{sinu}}}}{{\rm{k}}}} \right)\left(
{{\rm{h\: }}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\:
}}0\frac{{\rm{k}}}{{\rm{h}}}} \right)$
= k $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2\cos
\frac{{{\rm{u}} + {\rm{k}} + {\rm{u}}}}{2}.\sin \frac{{{\rm{u}} + {\rm{k}} -
{\rm{u}}}}{2}}}{{\rm{k}}}$ * ${\rm{h\:
}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\: }}0$$\frac{{\log
\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{logx}}}}{{\rm{h}}}$.
= k$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2\cos
\left( {{\rm{u}} + \frac{{\rm{k}}}{2}} \right).\sin
\frac{{\rm{k}}}{2}}}{{\frac{{\rm{k}}}{2}.2}}$. ${\rm{h\:
}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\: }}0$$\frac{{\log
\left( {\frac{{{\rm{x}} + {\rm{h}}}}{{\rm{x}}}} \right)}}{{\rm{h}}}$
= cosu.1. ${\rm{h\:
}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\: }}0$$\frac{{\log
\left( {1 + \frac{{\rm{h}}}{{\rm{x}}}}
\right)}}{{\frac{{\rm{h}}}{{\rm{x}}}.{\rm{x}}}}$ = cosu.1.$\frac{1}{{\rm{x}}}$
= $\frac{1}{{\rm{x}}}$.cos(logx).
(iii) $\sqrt {\tan x} $
Solution:
f(x) = $\sqrt {{\rm{tanx}}} $ = x.logx , f(x + h) = $\sqrt {\tan \left(
{{\rm{x}} + {\rm{h}}} \right)} $.
f’(x) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0$\frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left(
{\rm{x}} \right)}}{{\rm{h}}}$
or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\sqrt {{\rm{tanx}}} } \right)$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sqrt
{{\rm{tan}}\left( {{\rm{x}} + {\rm{h}}} \right)} ---\sqrt {{\rm{tanx}}}
}}{{\rm{h}}}$ ….(i)
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sqrt
{\tan \left( {{\rm{x}} + {\rm{h}}} \right)} - \sqrt {{\rm{tanx}}}
}}{{\rm{h}}}.\frac{{\sqrt {\tan \left( {{\rm{x}} + {\rm{h}}} \right)} +
\sqrt {{\rm{tanx}}} }}{{\sqrt {\tan \left( {{\rm{x}} + {\rm{h}}}
\right)} + \sqrt {{\rm{tanx}}} }}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\tan
\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{tanx}}}}{{\sqrt {\tan \left(
{{\rm{x}} + {\rm{h}}} \right)} + \sqrt {{\rm{tanx}}} }}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{\frac{{\sin \left( {{\rm{x}} + {\rm{h}}} \right)}}{{\cos \left(
{{\rm{x}} + {\rm{h}}} \right)}} -
\frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}}}{{{\rm{h}}\left( {\sqrt {\tan \left(
{{\rm{x}} + {\rm{h}}} \right)} + \sqrt {{\rm{tanx}}} } \right)}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sin
\left( {{\rm{x}} + {\rm{h}}} \right).{\rm{cosx}} - \cos \left( {{\rm{x}} +
{\rm{h}}} \right).{\rm{sinx}}}}{{{\rm{h}}.\cos \left( {{\rm{x}} + {\rm{h}}}
\right).\cos \left( {\sqrt {\tan \left( {{\rm{x}} + {\rm{h}}} \right)} +
\sqrt {{\rm{tanx}}} } \right)}}$.
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sin
\left( {{\rm{x}} + {\rm{h}} - {\rm{x}}} \right)}}{{{\rm{h}}.\cos \left(
{{\rm{x}} + {\rm{h}}} \right).{\rm{cosx}}\left( {\sqrt {\tan \left( {{\rm{x}}
+ {\rm{h}}} \right)} + \sqrt {{\rm{tanx}}} } \right)}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{sinh}}}}{{{\rm{hcos}}\left( {{\rm{x}} + {\rm{h}}}
\right).{\rm{cosx}}\left( {\sqrt {\tan \left( {{\rm{x}} + {\rm{h}}}
\right)} + \sqrt {{\rm{tanx\: }}} } \right)}}$.
= $\frac{1}{{{\rm{cosx}}.{\rm{cosx}}\left( {\sqrt {{\rm{tan\: x}}} +
\sqrt {{\rm{tanx}}} } \right)}}$ = $\frac{{{{\sec }^2}{\rm{x}}}}{{2\sqrt
{{\rm{tanx}}} }}$.
6. Find, from first principles, the derivatives of:
(i) f(x) = ecosx at x=0
Solution:
f(x) = ecosx
f’(0) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{f}}\left( {0 + {\rm{h}}} \right) - {\rm{f}}\left( 0
\right)}}{{\rm{h}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{{{\rm{e}}^{\cos \left( {0 + {\rm{h}}} \right)}} -
{{\rm{e}}^{{\rm{cos}}0}}}}{{\rm{h}}}$.
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\rm{e}}^{{\rm{cosh}}}} - {\rm{e}}}}{{\rm{h}}}$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{e}}\left(
{{{\rm{e}}^{{\rm{cosh}} - 1}} - 1} \right)}}{{\rm{h}}}$ …(i).
Put cosh – 1 = u. When h → 0, u → 0.
Then (i) becomes,
Or, f’(0) = u $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{e}}\left({{{\rm{e}}^{\rm{u}}} - 1} \right)}}{{\rm{h}}}$ = u
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
${\rm{e}}.\frac{{{{\rm{e}}^{\rm{u}}} -
1}}{{\rm{u}}}.\frac{{\rm{u}}}{{\rm{h}}}$.
= e $\left( {{\rm{u\:
}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\:
}}0\frac{{{{\rm{e}}^{\rm{u}}} - 1}}{{\rm{u}}}} \right)\left( {{\rm{h\:
}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\ \to\end{array}{\rm{\:
}}0\frac{{\rm{u}}}{{\rm{h}}}} \right)$ = e.1. h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\ \to\end{array}$ 0
$\frac{{{\rm{cosh}} - 1}}{{\rm{h}}}$.
= e. h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{ -
2{{\sin }^2}\frac{{\rm{h}}}{2}}}{{\rm{h}}}$ = -2e h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 ${\left( {\frac{{\sin
\frac{{\rm{h}}}{2}}}{{\frac{{\rm{h}}}{2}}}} \right)^2}.\frac{{\rm{h}}}{4}$ =
0.
(ii) f(x) = log.cosx at x=0
Solution:
f(x) = log.cosx
f’(0) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{f}}\left( {0 + {\rm{h}}} \right) - {\rm{f}}\left( 0
\right)}}{{\rm{h}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{{\rm{log}}.{\rm{cosh}} - 0}}{{\rm{h}}}$.
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{log}}.{\rm{cosh}}}}{{\rm{h}}}$ = h
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{log}}(1 -
2{{\sin }^2}\frac{{2{\rm{h}}}}{2}){\rm{\: }}}}{{\rm{h}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{1}{{\rm{h}}}$$\left\{ { - 2{{\sin }^2}\frac{{\rm{h}}}{2} - {{\left(
{2{{\sin }^2}\frac{{\rm{h}}}{2}} \right)}^2} - \ldots } \right\}$ =
0.
7. (i) $if\;{\rm{ f(x) = }}\left\{ {\begin{array}{*{20}{c}}{{x^2}\cos
\frac{1}{x}{\rm{ for x}} \ne {\rm{0}}}\\{0{\rm{ \;for \;x = 0}}}\end{array}}
\right..{\rm{ Find\; f'(0)}}$
Solution:
Rf’(0) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{f}}\left( {0 + {\rm{h}}} \right) - {\rm{f}}\left( 0
\right)}}{{\rm{h}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{{{\rm{h}}^2}\cos \frac{1}{{\rm{h}}} - 0}}{{\rm{h}}}$.
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 h.cos
$\frac{1}{{\rm{h}}}$ = 0.
Lf’ (0) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{f}}\left( {0 - {\rm{h}}} \right) - {\rm{f}}\left( 0 \right)}}{{ -
{\rm{h}}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{{\left( { - {\rm{h}}} \right)}^2}.\cos \left( { - \frac{1}{{\rm{h}}}}
\right) - 0}}{{ - {\rm{h}}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $ - {\rm{h}}.\cos
\frac{1}{{\rm{h}}}$ = 0
LF’(0) = Rf’(0)
So, f(0) exists and is equal to 0.
(ii)
$If\;{\rm{ f(x) = }}\left\{ {\begin{array}{*{20}{c}}{3x + 2{\rm{
\;for\; - }}\frac{3}{2} \le x \le 0}\\{3x - 2{\rm{ \;for \;0 < x
< }}\frac{3}{2}}\end{array}} \right.$
Show that f(x) is continuous at x=0, but f’(0) does not exist.
Solution
Or, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – f(x) = x
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – (3 + 2x) = 3.
Or, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + f(x) = x
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + (3 – 2x) = 3.
When x = 0, f(0) = 3 + 2 * 0 = 3.
x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – f(x) = x
$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + f(x) = f(0), so f(x)
is continuous at x = 0.
Lf’(0) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{f}}\left( {0 - {\rm{h}}} \right) - {\rm{f}}\left( 0 \right)}}{{ -
{\rm{h}}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{3 + 2\left( { - {\rm{h}}} \right) - 3}}{{ - {\rm{h}}}}$ = 2.
Rf’(0) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{f}}\left( {0 + {\rm{h}}} \right) - {\rm{f}}\left( 0
\right)}}{{\rm{h}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{3 - 2{\rm{h}} - 3}}{{\rm{h}}}$ = -2.
So, Lf’(0) ≠ Rf’(0)
So, f’(0) does not exist.
(iii) If f(x) =|x|, show that f’(0) does not exist.
Solution
f(x) = |x|;f(x) = x if x > 0, f(x) = - x if x < 0.
Lf’(0) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{f}}\left( {0 - {\rm{h}}} \right) - {\rm{f}}\left( 0 \right)}}{{ -
{\rm{h}}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\ \to\end{array}$ 0
$\frac{{--\left( { - {\rm{h}}} \right) - 0}}{{ - {\rm{h}}}}$ = -1.
Rf’(0) = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0
$\frac{{{\rm{f}}\left( {0 + {\rm{h}}} \right) - {\rm{f}}\left( 0
\right)}}{{\rm{h}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$
0 $\frac{{{\rm{h}} - 0}}{{\rm{h}}}$ = 1.
So, Lf’(0) ≠ Rf’(0)
So, f’(0) does not exist.
8) i. y = $\frac{{{\rm{x}}\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{2} -
\frac{{{{\rm{a}}^2}}}{2}\log \left( {{\rm{x}} + \sqrt {{{\rm{x}}^2} -
{{\rm{a}}^2}} } \right)$
Solution:
y = $\frac{{{\rm{x}}\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{2} -
\frac{{{{\rm{a}}^2}}}{2}\log \left( {{\rm{x}} + \sqrt {{{\rm{x}}^2} -
{{\rm{a}}^2}} } \right)$
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{1.\sqrt {{{\rm{x}}^2} -
{{\rm{a}}^2}} }}{2}$ + x. $\frac{1}{{2\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}}
}}.\frac{{2{\rm{x}}}}{2}$ – $\frac{{{{\rm{a}}^2}\left( {1 + \frac{1}{{2\sqrt
{{{\rm{x}}^2} - {{\rm{a}}^2}} }}.2{\rm{x}}} \right)}}{{2\left( {{\rm{x}}\sqrt
{{{\rm{x}}^2} - {{\rm{a}}^2}} } \right)}}$
= $\frac{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{2} +
\frac{{{{\rm{x}}^2}}}{{2\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }} -
\frac{{{{\rm{a}}^2}}}{2}\frac{{\left( {\sqrt {{{\rm{x}}^2} -
{{\rm{a}}^2}} + {\rm{x}}} \right)}}{{\left( {{\rm{x}} + \sqrt
{{{\rm{x}}^2} - {{\rm{a}}^2}} } \right)\left( {\sqrt {{{\rm{x}}^2} -
{{\rm{a}}^2}} } \right)}}{\rm{\: }}$
= $\frac{1}{2}$$\left( {\frac{{{{\rm{x}}^2} - {{\rm{a}}^2} + {{\rm{x}}^2} -
{{\rm{a}}^2}}}{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}} \right)$ =
$\frac{1}{2}.\frac{{2{{\rm{x}}^2} - 2{{\rm{a}}^2}}}{{\sqrt {{{\rm{x}}^2} -
{{\rm{a}}^2}} }}$.
= $\frac{{{{\rm{x}}^2} - {{\rm{a}}^2}}}{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}}
}}$ = $\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} $.
(ii)$y = \frac{{x\sqrt {{a^2} - {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sin ^{ -
1}}\frac{x}{a}$
Solution:
$\begin{array}{l}\frac{d}{{dx}}\left[ {\frac{{x\sqrt {{a^2} - {x^2}} }}{2} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{x}{a}} \right]\\ = \frac{d}{{dx}}\left[ {\frac{{x\sqrt {{a^2} - {x^2}} }}{2}} \right] + \frac{d}{{dx}}\left[ {\frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{x}{a}} \right]\\ = \frac{x}{2}\frac{d}{{dx}}\left[ {\sqrt {{a^2} - {x^2}} } \right] + \left[ {\sqrt {{a^2} - {x^2}} } \right]\frac{d}{{dx}}\frac{x}{2} + \frac{{{a^2}}}{2}\frac{d}{{dx}}{\sin ^{ - 1}}\frac{x}{a}\\ = \frac{x}{2}\frac{d}{{dx}}\left[ {\sqrt {{a^2} - {x^2}} } \right] + \left[ {\sqrt {{a^2} - {x^2}} } \right]\frac{d}{{dx}}\frac{x}{2} + \frac{{{a^2}}}{2}.\frac{1}{{\sqrt {1 - {{\left( {\frac{x}{a}} \right)}^2}} }}.\frac{1}{a}\\ = - \frac{{{x^2}}}{2}.\frac{1}{{\sqrt {{a^2} - {x^2}} }} + \frac{1}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{{2\sqrt {{a^2} - {x^2}} }}\\ = \frac{1}{{2\sqrt {{a^2} - {x^2}} }}({a^2} - {x^2}) + \frac{1}{2}\sqrt {{a^2} - {x^2}} \\ = \frac{{\sqrt {{a^2} - {x^2}} }}{2} + \frac{1}{2}\sqrt {{a^2}{x^2}} \\ = \sqrt {{a^2} - {x^2}} \end{array}$
(iii) y = 2tan-1$\frac{{\sqrt {1 + {{\rm{x}}^2}} -
1}}{{\rm{x}}}$
Solution:
y = 2tan-1$\frac{{\sqrt {1 + {{\rm{x}}^2}} - 1}}{{\rm{x}}}$
= 2tan-1$\left( {\frac{{\sin \theta - 1}}{{{\rm{tan}}\theta
}}} \right)$ [x = tanθ]
= 2tan-1. $\left( {\frac{{1 - {\rm{cos}}\theta }}{{{\rm{sin}}\theta
}}} \right)$ = 2tan-1. Tan $\frac{\theta }{2}$ = 2.$\frac{\theta
}{2}$ = θ.
So, y = tan-1x [x = tanθ]
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{{1 + {{\rm{x}}^2}}}$.
9) i. If xy = ex – y, prove that
$\frac{{dy}}{{dx}}$ = (log x) (log(ex)-2
Solution
xy = ex – y
or, log xy = log ex – y,
So, y.logx = x – y [log e = 1]
Or, y(logx + 1) = x
So, y = $\frac{{\rm{x}}}{{1 + {\rm{logx}}}}$.
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\left( {1 + {\rm{logx}}}
\right).1 - {\rm{x}}.\left( {0 + \frac{1}{{\rm{x}}}} \right)}}{{{{\left( {1 +
{\rm{logx}}} \right)}^2}}}$.
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{logx}}}}{{{{\left( {1 +
{\rm{logx}}} \right)}^2}}}$ = $\frac{{{\rm{logx}}}}{{{{\left( {{\rm{loge}} +
{\rm{logx}}} \right)}^2}}}$ = $\frac{{{\rm{logx}}}}{{{{\left( {{\rm{logex}}}
\right)}^2}}}$.
= logx.(log ex)-2.
(ii) If sin y = x sin (a+y), prove that $\frac{{dy}}{{dx}}$ =
$\frac{{{{{\mathop{\rm Sin}\nolimits} }^2}(a + y)}}{{\sin a}}$
Solution
Siny = x.sin(a + y) …(i)
Or, cosy. $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 1.sin(a + y) + x.cosx(a + y).
$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\sin \left( {{\rm{a}} +
{\rm{y}}} \right)}}{{{\rm{cosy}} - {\rm{x}}.\cos \left( {{\rm{a}} + {\rm{y}}}
\right)}}$
= $\frac{{\sin \left( {{\rm{a}} + {\rm{y}}} \right)}}{{{\rm{cosy}} -
\frac{{{\rm{siny}}}}{{\sin \left( {{\rm{a}} + {\rm{y}}} \right)}}.\cos \left(
{{\rm{a}} + {\rm{y}}} \right)}}$ [from (i)]
= $\frac{{{{\sin }^2}\left( {{\rm{a}} + {\rm{y}}} \right)}}{{\sin \left(
{{\rm{a}} + {\rm{y}}} \right).{\rm{cosy}} - {\rm{siny}}.\cos \left( {{\rm{a}}
+ {\rm{y}}} \right)}}$ = $\frac{{{{\sin }^2}\left( {{\rm{a}} + {\rm{y}}}
\right)}}{{\sin \left( {{\rm{a}} + {\rm{y}} - {\rm{y}}} \right)}}$ =
$\frac{{{{\sin }^2}\left( {{\rm{a}} + {\rm{y}}} \right)}}{{{\rm{sina}}}}$.
10) a. xsinx
Solution
Let y = xsinx.
So, logy = logxsinx = sinx.logx
So, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =
sinx.$\frac{1}{{\rm{x}}}$ + logx.cosx
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = xsinx$\left(
{\frac{{{\rm{sinx}}}}{{\rm{x}}} + {\rm{cosx}}.{\rm{logx}}} \right)$.
b. (sinx)x
Solution:
y = (sinx)x
So, logy = log(sinx)x = x.log(sinx)
So, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 1.log(sinx)
+ x.$\frac{1}{{\sin x}}$.cosx
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = (sinx)x(log.sinx +
x.cotx).
c. (sinx)logx
Solution:
y = (sinx)logx
So, logy = log(sinx)logx = logx.log(sinx)
So, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =
logx.$\frac{1}{{{\rm{sinx}}}}$.cosx + $\frac{1}{{\rm{x}}}$.log(sinx)
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = (sinx)logx$\left(
{{\rm{cotx}}.{\rm{logx}} + \frac{1}{{\rm{x}}}.{\rm{logsinx}}} \right)$.
d. ${{\rm{e}}^{{{\rm{x}}^{\rm{x}}}}}$
Solution:
y = ${{\rm{e}}^{{{\rm{x}}^{\rm{x}}}}}$
So, y = et.
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = et.
Where, t = xx.
Or, logt = x.logx
So, $\frac{1}{{\rm{t}}}$.$\frac{{{\rm{dt}}}}{{{\rm{dx}}}}$ =
x.$\frac{1}{{\rm{x}}}$ + 1.logx
So, $\frac{{{\rm{dt}}}}{{{\rm{dx}}}}$ = xx(1 + logx).
We have, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =
$\frac{{{\rm{dy}}}}{{{\rm{dt}}}}.\frac{{{\rm{dt}}}}{{{\rm{dx}}}}$ =
et.xx(1 + logx) =
${{\rm{e}}^{{{\rm{x}}^{\rm{x}}}}}$.xx(1 + logx).
e. ${{\rm{x}}^{{{\rm{e}}^{\rm{x}}}}}$
Solution:
y = ${{\rm{x}}^{{{\rm{e}}^{\rm{x}}}}}$
So, logy = ${\rm{log}}{{\rm{x}}^{{{\rm{e}}^{\rm{x}}}}}$=
ex.logx.
So, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =
ex.$\frac{1}{{\rm{x}}}$ + ex.logx
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =
${{\rm{x}}^{{{\rm{e}}^{\rm{x}}}}}\left(
{\frac{{{{\rm{e}}^{\rm{x}}}}}{{\rm{x}}} + {{\rm{e}}^{\rm{x}}}{\rm{logx}}}
\right)$ = ${{\rm{x}}^{{{\rm{e}}^{\rm{x}}}}}$.ex.$\left(
{{\rm{logx}} + \frac{1}{{\rm{x}}}} \right)$.
f. (logx)tanx
Solution:
y = (logx)tanx
So, logy = log(logx)tanx = tanx.log(logx)
So, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =
sec2x.log(logx) +
tanx.$\frac{1}{{{\rm{logx}}}}.\frac{1}{{\rm{x}}}$.
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = (log)tanx$\left( {{{\sec
}^2}{\rm{x}}.\log \left( {{\rm{logx}}} \right) +
\frac{{{\rm{tanx}}}}{{{\rm{x}}.{\rm{logx}}}}} \right)$.
g. xsecx
Solution
y = xsecx
So, logy = logxsecx = secx.logx
So, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = secx.tanx +
secx.$\frac{1}{{\rm{x}}}$.
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = xsecx$\left(
{{\rm{secx}}.{\rm{tanx}}.{\rm{logx}} + \frac{{{\rm{secx}}}}{{\rm{x}}}}
\right)$.
h. (sinx)cosx
Solution:
y = (sinx)cosx
So, logy = logx(sinx)cosx = cosx.log(sinx)
So, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = -sinx.log(sinx) =
cosx.$\frac{1}{{{\rm{sinx}}}}$.cosx
= (sinx)cosx(cosx.cotx – sinx.logsinx).
11) a. xx.yx=1
Solution:
xy.yx = 1.
Or, log(xy.yx) = log 1.
Or, log xy + log yx = 0
So, y.logx + x.logy = 0
Or, y. $\frac{1}{{\rm{x}}}$ + $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.logx + 1.logy
+ x.$\frac{1}{{\rm{y}}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0
Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{ - \frac{{\rm{y}}}{{\rm{x}}} -
{\rm{logy}}}}{{{\rm{logx}} + \frac{{\rm{x}}}{{\rm{y}}}}}$ = $ -
\frac{{\rm{y}}}{{\rm{x}}}\left( {\frac{{{\rm{y}} + {\rm{xlogy}}}}{{{\rm{x}} +
{\rm{ylogx}}}}} \right)$.
b. xm.yn=(x+y)m+n
Solution:
or, xm.yn = (x + y)m + n …(I)
or, xm.nyn – 1. $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ +
mxm – 1.yn = (m + n)(x + y)m + n – 1.$\left( {1 + \frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right){\rm{\: }}.$
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\left( {{\rm{m}} + {\rm{n}}}
\right).{{\left( {{\rm{x}} + {\rm{y}}} \right)}^{{\rm{m}} + {\rm{n}} - 1}} -
{\rm{m}}{{\rm{x}}^{{\rm{m}} -
1}}.{{\rm{y}}^{\rm{n}}}}}{{{\rm{n}}{{\rm{x}}^{\rm{m}}}{{\rm{y}}^{{\rm{n}} -
1}} - \left( {{\rm{m}} + {\rm{n}}} \right){{\left( {{\rm{x}} + {\rm{y}}}
\right)}^{{\rm{m}} + {\rm{n}} - 1}}}}$
= $\frac{{\left( {{\rm{m}} + {\rm{n}}} \right).{{\left( {{\rm{x}} + {\rm{y}}}
\right)}^{{\rm{m}} + {\rm{n}}}} - {\rm{m}}{{\rm{x}}^{{\rm{m}} -
1}}.{{\rm{y}}^{\rm{n}}}\left( {{\rm{x}} + {\rm{y}}}
\right)}}{{{\rm{n}}{{\rm{x}}^{\rm{m}}}{{\rm{y}}^{{\rm{n}} - 1\left( {{\rm{x}}
+ {\rm{y}}} \right)}} - \left( {{\rm{m}} + {\rm{n}}} \right){{\left( {{\rm{x}}
+ {\rm{y}}} \right)}^{{\rm{m}} + {\rm{n}}}}}}$
[multiplying by (x+y)]
Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\left( {{\rm{m}} + {\rm{n}}}
\right).{{\rm{x}}^{\rm{m}}}{{\rm{y}}^{\rm{n}}} - {\rm{m}}{{\rm{x}}^{{\rm{m}} -
1}}.{{\rm{y}}^{\rm{n}}}\left( {{\rm{x}} + {\rm{y}}}
\right)}}{{{{\rm{x}}^{\rm{m}}}{{\rm{y}}^{{\rm{n}} - 1}}\left\{ {{\rm{n}}\left(
{{\rm{x}} + {\rm{y}}} \right) - \left( {{\rm{m}} + {\rm{n}}} \right){\rm{y}}}
\right\}}}$
= $\frac{{\rm{y}}}{{\rm{x}}}$$\left( {\frac{{{\rm{mx}} + {\rm{nx}} - {\rm{mx}}
- {\rm{my}}}}{{{\rm{nx}} + {\rm{ny}} - {\rm{my}} - {\rm{ny}}}}} \right)$ =
$\frac{{\rm{y}}}{{\rm{x}}}$.
c. esinx + esiny = 1
Solution:
esinx + esiny = 1
Or, esinx.cosx + esiny.coy
$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0 [on diff]
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{ -
{{\rm{e}}^{{\rm{sinx}}}}.{\rm{cosx}}}}{{{{\rm{e}}^{{\rm{siny}}}}.{\rm{cosy}}}}$.
d) xy = yx
Solution:
or, xy = yx
or, log xy = log yx [taking log on both
sides]
or, y.logx = x.logy
or, y.$\frac{1}{{\rm{x}}}$ + logx. $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =
x.$\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + 1.logy
[on differ.]
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\log {\rm{y}} -
\frac{{\rm{y}}}{{\rm{x}}}}}{{\log {\rm{x}} - \frac{{\rm{x}}}{{\rm{y}}}}}$ =
$\frac{{\rm{y}}}{{\rm{x}}}\left( {\frac{{{\rm{xlogy}} -
{\rm{y}}}}{{{\rm{ylogx}} - {\rm{x}}}}} \right)$.
e. xsinx = ysiny
Solution:
xsinx = ysiny
Or, sinx.logx = siny.logy [taking log on both sides]
Or, cosx.logx + sinx. $\frac{1}{{\rm{x}}}$ = cosy.
$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.logy +
siny$.\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{y}}}{{\rm{x}}}\left(
{\frac{{{\rm{x}}.{\rm{cosx}}.{\rm{logx}} +
{\rm{sinx}}}}{{{\rm{y}}.{\rm{cosy}}.{\rm{logy}} + {\rm{siny}}}}} \right)$.
12. Find the derivative of:
(i) xtanx + (tanx)x
Solution
Let, y = xtanx + (tanx)x
Or, y = u + v →$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =
$\frac{{{\rm{du}}}}{{{\rm{dx}}}} + \frac{{{\rm{dv}}}}{{{\rm{dx}}}}$ ….(i)
Where, u = xtanx and v = (tanx)x
Log u = tanx.logx
Logv = xlog(tanx).
So, $\frac{1}{{\rm{u}}}.\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ =
sec2x.logx + tanx.$\frac{1}{{\rm{x}}}$
So, $\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ = xtanx(sec2x.logx
$\frac{{{\rm{tanx}}}}{{\rm{x}}}$)
So, $\frac{1}{{\rm{v}}}.\frac{{{\rm{dv}}}}{{{\rm{dx}}}}$ = 1.logtanx +
$\frac{{\rm{x}}}{{{\rm{tanx}}}}$.sec2x
So, $\frac{{{\rm{dv}}}}{{{\rm{dx}}}}$ = (tanx)x$\left(
{{\rm{logtanx}} + \frac{{{\rm{xse}}{{\rm{c}}^2}{\rm{x}}}}{{{\rm{tanx}}}}}
\right)$.
Thus, from (i),
Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = xtanx$\left( {{{\sec
}^2}{\rm{xlogx}} + \frac{{{\rm{tanx}}}}{{\rm{x}}}} \right)$ +
(tanx)x(log.tanx + 2x.cosec2x)
(ii) (tanx)cotx + (cotx)tanx
Solution:
Let y = (tanx)cotx + (cotx)tanx
So, y = u + v →$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =
$\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ + $\frac{{{\rm{dv}}}}{{{\rm{dx}}}}$ …(i)
Where, u = (tanx)cotx and v = (cotx)tanx.
Since, u = (tanx)cotx.
So, logu = cotx.log(tanx) [on taking log]
Or, $\frac{1}{{\rm{u}}}$.$\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ = cotx.
$\frac{1}{{{\rm{tanx}}}}$.sec2x – cosec2x.log(tanx)
So, $\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ =
(tanx)cotx{cosec2x(1 – logtanx)}.
Again, v = (cotx)tanx.
Or, logv = tanx.log(cotx)
Or, $\frac{1}{{\rm{v}}}.\frac{{{\rm{dv}}}}{{{\rm{dx}}}}$ =
sec2xlog(cotx) +
tanx.$\frac{1}{{{\rm{cotx}}}}$.(-cosec2x).
So, $\frac{{{\rm{dv}}}}{{{\rm{dx}}}}$ =
(cotx)tanx(sec2x.logcotx – sec2x)
So, $\frac{{{\rm{dv}}}}{{{\rm{dx}}}}$ =
(cotx)tanx{sec2x(log.cotx – 1)}
Thus from (i),
Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =
(tanx)cotx{cosec2x(1 – logtanx) +
(cotx)tanx{sec2x(log.cotx – 1)}