Question

Differentiate the given functions w.r.t. x.

$(xcosx{)}^x+(xsinx{)}^{\frac{1}{x}}.$

Answer 1

Let y = $(xcosx{)}^x+(xsinx{)}^{\frac{1}{x}}$

Let $u=(xcosx{)}^x,v=(xsinx{)}^{\frac{1}{x}}$

$\therefore$ y = u + v

Differentiating w.r.t. x, we get $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

Now, $u=(xcosx{)}^x$

Taking log on both sides, log u = x log (x cos x)

Differentiating w.r.t. x,

$\frac{d}{dx}\left(logu\right)=x\frac{d}{dx}log\left(xcosx\right)+log\left(xcosx\right)\frac{d}{dx}\left(x\right)\Rightarrow \frac{1}{u}\frac{du}{dx}=x\times \frac{1}{xcosx}\frac{d}{dx}\left(xcosx\right)+log\left(xcosx\right)\times 1\Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{1}{cosx}(x\frac{d}{dx}cosx+cosx\frac{d}{dx}x)+log\left(xcosx\right)\Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{1}{cosx}\left[x\right(-sinx)+cosx]+log\left(xcosx\right)=\frac{1}{cosx}[-xsinx+cosx]+log\left(xcosx\right)\Rightarrow \frac{1}{u}\frac{du}{dx}=-xtanx+1+log\left(xcosx\right)\Rightarrow \frac{du}{dx}=u[-xtanx+1+log(xcosx\left)\right]\Rightarrow \frac{du}{dx}=\left(xcosx{)}^x\right[-xtanx+1+log\left(xcosx\right)]Now,v=(xsinx{)}^{1/x}Takinglogonbothsides,logv=\frac{1}{x}log\left(xsinx\right)Differentiatingw.r.t.x,\frac{1}{v}\frac{d}{dx}logv=\frac{1}{x}\frac{d}{dx}\left(logxsinx\right)+log\left(xsinx\right)\frac{d}{dx}\left(\frac{1}{x}\right)\frac{1}{v}\frac{dv}{dx}=\frac{1}{x}\left[\frac{1}{xsinx}\frac{d}{dx}\right(xsinx\left)\right]+log\left(xsinx\right)(-\frac{1}{x^2})=\frac{1}{x}\left[\frac{1}{xsinx}\{x\frac{d}{dx}sinx+sinx\frac{d}{dx}x\}\right]-\left(\frac{1}{x^2}\right)log\left(xsinx\right)\Rightarrow \frac{1}{x}\frac{dv}{dx}=\frac{1}{x}\frac{1}{xsinx}[xcosx+sinx]+log\left(xsinx\right)\left(\frac{-1}{x^2}\right)=\frac{xcosx}{x^{2}sinx}+\frac{sinx}{x^{2}sinx}+\left(\frac{-1}{x^2}\right)logxsinx\Rightarrow \frac{dv}{dx}=v[\frac{cotx}{x}+\frac{1}{x^2}-\frac{1}{x^2}logxsinx]\Rightarrow \frac{dv}{dx}=v\left[\frac{xcotx+1-logxsinx}{x^2}\right]\Rightarrow \frac{dv}{dx}=(xsinx{)}^{1/x}\left[\frac{xcotx+1-logxsinx}{x^2}\right]Now,puttingthevaluesof\frac{du}{dx}and\frac{dv}{dx}inEq.(i)\frac{dy}{dx}=(xcosx{)}^x[-xtanx+1+log(xcosx\left)\right]+(xsinx{)}^{1/x}\left[\frac{xcotx+1-logxsinx}{x^2}\right]$