Continuity and Differentiability — Maths STD 12 Science — Question

Putting u = (x cos x)^x and $v = {\left( {x\sin x} \right)^{\frac{1}{x}}}$

we have y = u + v

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

Now u = (x cos x)^x

$\Rightarrow \log u = \log {\left( {x\cos x} \right)^x} = x\log \left( {x\cos x} \right)$

$\Rightarrow \log u = x\left( {\log x + \log \cos x} \right)$

$\Rightarrow \frac{d}{{dx}}\log u = \frac{d}{{dx}}\left\{ {x\left( {\log x + \log \cos x} \right)} \right\}$

$\Rightarrow \frac{1}{u}\frac{{du}}{{dx}}$ $ = x\left[ {\frac{1}{x} + \frac{1}{{\cos x}}.\left( { - \sin x} \right)} \right] + \left( {\log x + \log \cos x} \right).1$

$\Rightarrow \frac{1}{u}\frac{{du}}{{dx}} $$ = \left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right]$

$\Rightarrow \frac{{du}}{{dx}} = u\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right]$

$\Rightarrow \frac{{du}}{{dx}} = {\left( {x\cos x} \right)^x}\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right]$ .....(ii)

Again $v = {\left( {x\sin x} \right)^{\frac{1}{x}}}$

$\Rightarrow \log v = \log {\left( {x\sin x} \right)^{\frac{1}{x}}} = \frac{1}{x}\log \left( {x\sin x} \right)$

$\Rightarrow \log v = \frac{1}{x}\left( {\log x + \log \sin x} \right)$

$\Rightarrow \frac{d}{{dx}}\log v = \frac{d}{{dx}}\left\{ {\frac{1}{x}\left( {\log x + \log \sin x} \right)} \right\}$

$\Rightarrow \frac{1}{v}\frac{dv}{{dx}} $$= \frac{1}{x}\left[ {\frac{1}{x} + \frac{1}{{\ sinx }}.\cos x} \right] + \left( {\log x + \log \sin x} \right)\left( {\frac{{ - 1}}{{{x^2}}}} \right)$

$\Rightarrow \frac{1}{v}\frac{dv}{{dx}} $$= \left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$

$\Rightarrow \frac{dv}{{dx}} = v\left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$

$\Rightarrow \frac{dv}{{dx}} = {\left( {x\sin x} \right)^{\frac{1}{x}}}\left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$ ...(iii)

Putting the values from eq. (ii) and (iii) in eq. (i)

$\frac{d}{{dx}} = {\left( {x\cos x} \right)^{^x}}$$\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right] + {\left( {x\sin x} \right)^{\frac{1}{x}}}$$\left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$