Derivatives Exercise: 15.2 Class 12 Basic Mathematics Solution [NEB UPDATED]

Syllabus: Content to study

Topic: Derivative of Hyperbolic Function

1. Hyperbolic Function
2. Derivative of Hyperbolic Function

Exercise 15.2

Derivative of Hyperbolic Function

1. log(tanh x)

Solution:

Let y = log(tanhx)

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}\left( {{\rm{logtanhx}}} \right)}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}\left( {{\rm{log}}.{\rm{tangx}}} \right)}}{{{\rm{d}}\left( {{\rm{tanhx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{tanhx}}} \right)}}{{{\rm{dx}}}}$.

= $\frac{1}{{{\rm{tanhx}}}}.{\sec ^2}{\rm{hx}}$ = $\frac{{{\rm{coshx}}}}{{{\rm{sinhx}}}}.\frac{1}{{{{\cosh }^2}{\rm{x}}}}$

= $\frac{1}{{{\rm{sinhx}}.{\rm{coshx}}}}$ = $\frac{2}{{{\rm{sinh}}2{\rm{x}}}}$ = 2cosec2x. 

2. log.sinh $\frac{{\rm{x}}}{{\rm{a}}}$

Solution:

Let y = log.sinh $\frac{{\rm{x}}}{{\rm{a}}}$

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{log}}.\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)$ = $\frac{{{\rm{d}}\left( {{\rm{log}}.\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{d}}\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {\sin \frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{dx}}}}$.

= $\frac{1}{{\sinh \frac{{\rm{x}}}{{\rm{a}}}}}$.$\cos \frac{{\rm{x}}}{{\rm{a}}}.\frac{1}{{\rm{a}}}$ = $\frac{1}{{\rm{a}}}.\coth \frac{{\rm{x}}}{{\rm{a}}}$. 

3. e^sinhx

Solution:

Let y = e^sinhx.

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(e^sinhx) = $\frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{sinhx}}}}} \right)}}{{{\rm{d}}\left( {{\rm{sinhx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{sinhx}}} \right)}}{{{\rm{dx}}}}$ = e^sinhx.coshx.

4. ${{\rm{e}}^{{{\cosh }^{ - 1}}\frac{{\rm{x}}}{{\rm{a}}}}}$

Solution:

Let y = ${{\rm{e}}^{{{\cosh }^{ - 1}}\frac{{\rm{x}}}{{\rm{a}}}}}$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}\left( {{{\rm{e}}^{{{\cos }^{ - 1}}\frac{{\rm{x}}}{{\rm{a}}}}}} \right)}}{{{\rm{d}}\left( {{{\cos }^{ - 1}}\frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {{{\cos }^{ - 1}}\frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{dx}}}}$.

= ${{\rm{e}}^{{{\cos }^{ - 1}}\frac{{\rm{x}}}{{\rm{a}}}}}$. $\frac{1}{{\sqrt {{{\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}^2} - 1} }}.\frac{1}{{\rm{a}}}$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{{\rm{e}}^{{{\cos }^{ - 1}}\frac{{\rm{x}}}{{\rm{a}}}}}}}{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}$. 

5. sech(tan^-1x)

Solution:

Let y = sech(tan^-1x)

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}\left( {{\rm{sech}}\left( {{{\tan }^{ - 1}}{\rm{x}}} \right)} \right)}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}\left( {{\rm{sech}}\left( {{{\tan }^{ - 1}}{\rm{x}}} \right)} \right)}}{{{\rm{d}}\left( {{{\tan }^{ - 1}}{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {{{\tan }^{ - 1}}{\rm{x}}} \right)}}{{{\rm{dx}}}}$

= - sech(tan^-1x).tanh(tan^-1x).$\frac{1}{{1 + {{\rm{x}}^2}}}$.

= $ - \frac{1}{{1 + {{\rm{x}}^2}}}$sech.(tan^-1x)tanh(tan^-1x).

6. sec^-1x – cos^-1x.

Solution:

Ley y = sec^-1x – cos^-1x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sech^-1x – cos^-1x)

= $\frac{{\rm{d}}}{{{\rm{dx}}}}$ (sech^-1x) – $\frac{{\rm{d}}}{{{\rm{dx}}}}$(cosh^-1x)

= $ - \frac{1}{{{\rm{x}}\sqrt {1 - {{\rm{x}}^2}} }} - \frac{1}{{\sqrt {{{\rm{x}}^2} - 1} }}$. 

7. Arc.tan.sinhx

Solution:

Let y = Arc.tan.sinhx

Or, y = tan^-1.sinhx

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(tan^-1.sinhx) = $\frac{{{\rm{d}}\left( {{{\tan }^{ - 1}}{\rm{sinhx}}} \right)}}{{{\rm{d}}\left( {{\rm{sinhx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{sinhx}}} \right)}}{{{\rm{dx}}}}$

= $\frac{1}{{1 + {{\sinh }^2}{\rm{x}}}}.{\rm{coshx}}$ = $\frac{{{\rm{coshx}}}}{{{{\cosh }^2}{\rm{x}}}}$ = sec.hx. 

8. 2tanh^-1$\left( {{\rm{tan}}\frac{1}{2}{\rm{x}}} \right)$

Solution:

Let y = 2tanh^-1$\left( {{\rm{tan}}\frac{1}{2}{\rm{x}}} \right)$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 2.$\frac{{\rm{d}}}{{{\rm{dx}}}}$$\left( {{{\tan }^{ - 1}}(\tan \frac{1}{2}{\rm{x}}){\rm{\: }}} \right)$

= $2\frac{{{\rm{d}}\left( {{{\tan }^{ - 1}}(\tan \frac{1}{2}{\rm{x}}){\rm{\: }}} \right)}}{{{\rm{d}}\left( {\tan \frac{1}{2}{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {\tan \frac{1}{2}{\rm{x}}} \right)}}{{{\rm{d}}\left( {\frac{1}{2}{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{1}{2}{\rm{x}}} \right)}}{{{\rm{dx}}}}$.

= $2.{\rm{\: }}\frac{1}{{1 - {{\tan }^2}\frac{1}{2}{\rm{x}}}}$.sec²$\frac{1}{2}$.x.$\frac{1}{2}$ = $\frac{{1 + {{\tan }^2}\frac{1}{2}{\rm{x}}}}{{1 - {{\tan }^2}\frac{1}{2}{\rm{x}}}}$.

= $\frac{1}{{{\rm{cosx}}}}$ = secx. 

9. x^{cosh x/a}

Solution:

Let y = x^{cosh x/a}

log y = cosh $\frac{{\rm{x}}}{{\rm{a}}}$.logx

or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(logy) = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\cosh \frac{{\rm{x}}}{{\rm{a}}}.{\rm{logx}}} \right)$.

or, $\frac{{{\rm{d}}\left( {{\rm{logy}}} \right)}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = cosh $\frac{{\rm{x}}}{{\rm{a}}}$.$\frac{{\rm{d}}}{{{\rm{dx}}}}$(logx) + logx $\frac{{{\rm{d}}\left( {\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{dx}}}}$.

or, $\frac{1}{{\rm{y}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = cosh${\rm{\: }}\frac{{\rm{x}}}{{\rm{a}}}$.$\frac{1}{{\rm{x}}}$ + logx.sinh $\frac{{\rm{x}}}{{\rm{a}}}$.$\frac{1}{{\rm{a}}}$.

or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = y$\left( {\frac{1}{{\rm{x}}}\cosh \frac{{\rm{x}}}{{\rm{a}}} + \frac{1}{{\rm{a}}}{\rm{logxsinh}}\frac{{\rm{x}}}{{\rm{a}}}} \right)$.

= x^{cosh x/a}.$\left( {\frac{1}{{\rm{x}}}\cosh \frac{{\rm{x}}}{{\rm{a}}} + \frac{1}{{\rm{a}}}{\rm{logx}}.\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)$. 

10. ${{\rm{x}}^{\sinh \frac{{{{\rm{x}}^2}}}{{\rm{a}}}}}$

Solution:

Let y = ${{\rm{x}}^{\sinh \frac{{{{\rm{x}}^2}}}{{\rm{a}}}}}$

So, logy = sinh $\frac{{{{\rm{x}}^2}}}{{\rm{a}}}$.logx

Differentiating, we have, $\frac{1}{{\rm{y}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = sinh${\rm{\: }}\frac{{{{\rm{x}}^2}}}{{\rm{a}}}.\frac{1}{{\rm{x}}}$ + cosh $\frac{{{{\rm{x}}^2}}}{{\rm{a}}}.\frac{{2{\rm{a}}}}{{\rm{a}}}$.logx.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = ${{\rm{x}}^{\sinh \frac{{{{\rm{x}}^2}}}{{\rm{a}}}}}$.${\rm{\: }}\left( {\frac{1}{{\rm{x}}}\sinh \frac{{{{\rm{x}}^2}}}{{\rm{a}}} + \frac{{2{\rm{x}}}}{{\rm{a}}}{\rm{logxcosh}}\frac{{{{\rm{x}}^2}}}{{\rm{a}}}} \right)$.

11. ${{\rm{x}}^{{{\cosh }^2}\frac{{\rm{x}}}{{\rm{a}}}}}$

Solution:

Let y = ${{\rm{x}}^{{{\cosh }^2}\frac{{\rm{x}}}{{\rm{a}}}}}$

log y = ${{\rm{x}}^{{{\cosh }^2}\frac{{\rm{x}}}{{\rm{a}}}}}$.logx

or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(logy) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$$\left( {{{\cosh }^2}\frac{{\rm{x}}}{{\rm{a}}}.{\rm{logx}}} \right)$.

or, $\frac{{{\rm{d}}\left( {{\rm{logy}}} \right)}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = cosh²$\frac{{\rm{x}}}{{\rm{a}}}$.$\frac{{\rm{d}}}{{{\rm{dx}}}}$(logx) + logx. $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\cosh }^2}\frac{{\rm{x}}}{{\rm{a}}}} \right)$.

Or, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = cosh²$\frac{{\rm{x}}}{{\rm{a}}}$.$\frac{1}{{\rm{x}}}$ + logx $\frac{{{\rm{d}}\left( {{{\cosh }^2}\frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{d}}\left( {\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{dx}}}}$.

= $\frac{1}{{\rm{x}}}$cosh²$\frac{{\rm{x}}}{{\rm{a}}}$ + log.x.2cosh $\frac{{\rm{x}}}{{\rm{a}}}$.sinh. $\frac{{\rm{x}}}{{\rm{a}}}$.$\frac{1}{{\rm{a}}}$.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = y$\left( {\frac{1}{{\rm{x}}}\cos {{\rm{h}}^2}{\rm{\: }}\frac{{\rm{x}}}{{\rm{a}}} + \frac{1}{{\rm{a}}}{\rm{logx}}.\sinh \frac{{2{\rm{x}}}}{{\rm{a}}}} \right)$

= ${{\rm{x}}^{{{\cosh }^2}\frac{{\rm{x}}}{{\rm{a}}}}}$. $\left( {\frac{1}{{\rm{x}}}{{\cosh }^2}\frac{{\rm{x}}}{{\rm{a}}} + \frac{1}{{\rm{a}}}{\rm{logxsinh}}\frac{{2{\rm{x}}}}{{\rm{a}}}} \right)$. 

12. ${\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)^{{{\rm{x}}^2}}}$

Solution:

Let y = ${\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)^{{{\rm{x}}^2}}}$

Or, logy = x²log.sinh $\frac{{\rm{x}}}{{\rm{a}}}$.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(log.y) = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}{\rm{log}}.\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)$.

Or, $\frac{{{\rm{d}}\left( {{\rm{logy}}} \right)}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = x².$\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{logsinh}}\frac{{\rm{x}}}{{\rm{a}}}} \right)$ + log.sinh $\frac{{\rm{x}}}{{\rm{a}}}$. $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x²).

Or, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = x²$\frac{{{\rm{d}}\left( {{\rm{log}}.\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{d}}\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{dx}}}}$ + log.sinh $\frac{{\rm{x}}}{{\rm{a}}}$.2x

Or, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = x².$\frac{1}{{\sinh \frac{{\rm{x}}}{{\rm{a}}}}}$.cosh $\frac{{\rm{x}}}{{\rm{a}}}$.$\frac{1}{{\rm{a}}}$ + 2x.log.sinh$\frac{{\rm{x}}}{{\rm{a}}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = y$.\left( {\frac{{{{\rm{x}}^2}}}{{\rm{a}}}.\coth \frac{{\rm{x}}}{{\rm{a}}} + 2{\rm{xlog}}.{\rm{sinh}}\frac{{\rm{x}}}{{\rm{a}}}} \right)$.

= ${\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)^{{{\rm{x}}^2}}}\left( {\frac{{{{\rm{x}}^2}}}{{\rm{a}}}.\coth \frac{{\rm{x}}}{{\rm{a}}} + 2{\rm{x}}.{\rm{log}}.\sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)$. 

13. ${\left( {\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)^{{\rm{logx}}}}$

Solution:

Let y = ${\left( {\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)^{{\rm{logx}}}}$

Or, logy = logx.log.cosh$\frac{{\rm{x}}}{{\rm{a}}}$.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(log y) = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{logx}}.{\rm{log}}.\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)$

Or, $\frac{{{\rm{d}}\left( {{\rm{logy}}} \right)}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = logx. $\frac{{{\rm{d}}\left( {{\rm{log}}.\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{dcosh}}\frac{{\rm{x}}}{{\rm{a}}}}}.\frac{{{\rm{d}}\left( {\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{dx}}}}$ + log.cosh$.\frac{{\rm{x}}}{{\rm{a}}}$.$\frac{{\rm{d}}}{{{\rm{dx}}}}$(logx).

Or, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = logx. $\frac{1}{{\cosh \frac{{\rm{x}}}{{\rm{a}}}}}$.sinh $\frac{{\rm{x}}}{{\rm{a}}}$.$\frac{1}{{\rm{a}}}$ + log.cosh $\frac{{\rm{x}}}{{\rm{a}}}$.$\frac{1}{{\rm{x}}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = y.$\left( {\frac{1}{{\rm{a}}}{\rm{logx}}.\tanh \frac{{\rm{x}}}{{\rm{a}}} + \frac{1}{{\rm{x}}}{\rm{log}}.\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)$.

= ${\left( {\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)^{{\rm{logx}}}}\left( {\frac{1}{{\rm{a}}}{\rm{logx}}.\tanh \frac{{\rm{x}}}{{\rm{a}}} + \frac{1}{{\rm{x}}}.{\rm{log}}.\cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)$. 

14. ${\left( {{\rm{coshx}}} \right)^{{{\sinh }^{ - 1}}{\rm{x}}}}$^.

Solution:

Let t = ${\left( {{\rm{coshx}}} \right)^{{{\sinh }^{ - 1}}{\rm{x}}}}$^.

So, logy = sinh^-1x.log(coshx)

So, $\frac{1}{{\rm{y}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = sinh^-1x.$\frac{1}{{{\rm{coshx}}}}$.sinhx + $\frac{1}{{\sqrt {1 + {{\rm{x}}^2}} }}$.log(coshx). 

15. ${\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)^{{\rm{nx}}}}$

Solution:

Let y = ${\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)^{{\rm{nx}}}}$

Or, logy = nx.log $\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)$

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(logy) = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left\{ {{\rm{nx}}.\log \left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)} \right\}$

Or, $\frac{{{\rm{d}}\left( {{\rm{logy}}} \right)}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = nx.$\frac{{\rm{d}}}{{{\rm{dx}}}}$log$\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)$ + log $\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)$.

Or, $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = nx.$\frac{{{\rm{d}}\left( {\log \left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)} \right)}}{{{\rm{d}}\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}$.$\frac{{{\rm{d}}\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{dx}}}}$ + log $(\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}){\rm{\: }}$.n.1

Or. $\frac{1}{{\rm{y}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = nx. $\frac{1}{{\sin \frac{{\rm{x}}}{{\rm{a}}} + \cos \frac{{\rm{x}}}{{\rm{a}}}}}$$\left( {\cosh \frac{{\rm{x}}}{{\rm{a}}} + \sinh \frac{{\rm{x}}}{{\rm{a}}}} \right)$.$\frac{1}{{\rm{a}}}$.$\frac{{{\rm{d}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)}}{{{\rm{dx}}}}$. + n.log$\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = y.$\left[ {\frac{{{\rm{nx}}}}{{\rm{a}}} + {\rm{n}}.\log \left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)} \right]$

= n.${\left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)^{{\rm{nx}}}}\left[ {\frac{{\rm{x}}}{{\rm{a}}} + \log \left( {\sinh \frac{{\rm{x}}}{{\rm{a}}} + \cosh \frac{{\rm{x}}}{{\rm{a}}}} \right)} \right]$.