Continuity and Differentiability — Maths STD 12 Science — Question

When $u = {x^{\sin x}}$, $v = {\left( {\sin x} \right)^{\cos x}}$

$\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}$ ...(i)

$u = {x^{\sin x}}$

Taking log both side

log $u = \log {x^{\sin x}}$

log $u = \sin x.\log x$

diff. both side w.r. to x

$\frac{1}{u}\frac{{du}}{{dx}} = \sin x.\frac{1}{x} + \log x\cos x$

$\frac{{du}}{{dx}} = u\left[ {\frac{{\sin x}}{x} + \log x.\cos x} \right]$

$\frac{{du}}{{dx}} = {x^{\sin x}}\left[ {\frac{{\sin x + x\log x.\cos x}}{x}} \right]$

$v = {\left( {\sin x} \right)^{\cos x}}$

Taking log both sides

log $v = \log {\left( {\sin x} \right)^{\cos x}}$

$\log v = \cos x.\log \left( {\sin x} \right)$

Differentiating both sides  w.r.t to x

$\frac{1}{v}.\frac{{dv}}{{dx}} = \cos x.\frac{1}{{\sin x}}\left( {\cos x} \right) + \log \left( {\sin x} \right)\left( { - \sin x} \right)$

$\frac{{dv}}{{dx}} = v\left[ {\cot x.\cos x - \log \left( {\sin x} \right).\sin x} \right]$

$\frac{{dv}}{{dx}} = {\left( {\sin x} \right)^{\cos x}}\left[ {\cot x.\cos x - \log \left( {\sin x} \right).\sin x} \right]$

Hence

$\frac{{dy}}{{dx}} = {x^{\sin x}}\left[ {\frac{{\sin x + x\log x.\cos x}}{x}} \right] + $${\left( {\sin x} \right)^{\cos x}}\left[ {\cot x\cos x - \log \left( {\sin x} \right).\sin x} \right]$