Question

Differentiate the function with respect to $x$
$(xcosx)^x+(xsinx)^{\frac{1}{x}}$

Answer 1

The correct answer is $\frac{dy}{dx}=(xcosx)^x[1-xtanx+log(xcosx)]+(xsinx)^{\frac{1}{x}}[\frac{xcotx+1-log(xsinx)}{x^2}]$
Let $y=(xcosx)^x+(xsinx)^{\frac{1}{x}}$
Also,let $u=(xcosx)^x$ and $v=(xsinx)^{\frac{1}{x}}$
$∴y=u+v$
$⇒\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} ....(1)$
$u=(xcosx)^x$
$⇒logu=log((xcosx)^x)$
$⇒logu=x(log(xcosx)$
$⇒logu=x[logx+logcosx]$
$⇒logu=xlogx+xlogcosx$
Differentiating both sides with respect to $x$,we obtain
$⇒\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(xlogx)+\frac{d}{dx}(xlogcosx)$
$⇒\frac{du}{dx}=u[{logx.\frac{d}{dx}(x)+x.\frac{d}{dx}(logx)}+{logcosx.\frac{d}{dx}(x)+x.\frac{d}{dx}(logcosx})]$
$⇒\frac{du}{dx}=(xcosx)^x[(logx.1+x.\frac{1}{x})+{logcosx.1+x.\frac{1}{cosx}.\frac{d}{dx}(cosx)}]$
$⇒\frac{du}{dx}=(xcosx)^x[(logx+1)+{logcosx+\frac{x}{cosx}.(-sinx)}]$
$⇒\frac{du}{dx}=(xcosx)^x[(1+logx)+logcosx-xtanx]$
$⇒\frac{du}{dx}=(xcosx)^x[1-xtanx+(logxlogcosx)]$
$⇒\frac{du}{dx}=(xcosx)^x[1-xtanx+log(xcosx)] ..(2)$
$v=(xsinx)^{\frac{1}{x}}$
$⇒logv=log((xsinx)^{\frac{1}{x}})$
$⇒logv=\frac{1}{x}log(xsinx)$
$⇒logv=\frac{1}{x}(logx+logsinx)$
$⇒logv=\frac{1}{x}logx+\frac{1}{x}logsinx$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}(\frac{1}{x}logx)+\frac{d}{dx}[(\frac{1}{x}log(sinx))]$
$⇒\frac{1}{v}\frac{dv}{dx}=[logx.\frac{d}{dx}(\frac{1}{x})+\frac{1}{x}.\frac{d}{dx}(logx)]+[log(sinx).\frac{d}{dx}(\frac{1}{x})+\frac{1}{x}.\frac{d}{dx}{log(sinx)}]$
$⇒\frac{1}{v}\frac{dv}{dx}=[logx.(\frac{-1}{x^2})+\frac{1}{x}.\frac{1}{x}]+[log(sinx).(\frac{-1}{x^2})+\frac{1}{x}.\frac{1}{sinx}.\frac{d}{dx}(sinx)]$
$⇒\frac{1}{v}\frac{dv}{dx}=\frac{1}{x^2}(1-logx)+[\frac{-log(sinx)}{x^2}+\frac{1}{xsinx}.cosx]$
$⇒\frac{1}{v}\frac{dv}{dx}=(xsinx)^{\frac{1}{x}}[\frac{1-logx}{x^2}+\frac{-log(sinx)+xcotx}{x^2}]$
$⇒\frac{1}{v}\frac{dv}{dx}=(xsinx)^{\frac{1}{x}}[\frac{1-logx-log(sinx)+xcotx}{x^2}]$
$⇒\frac{1}{v}\frac{dv}{dx}=(xsinx)^{\frac{1}{x}}[\frac{1-log(xsinx)+xcotx}{x^2}] ...(3)$
Therefore,from (1),(2),and(3),we obtain
$\frac{dy}{dx}=(xcosx)^x[1-xtanx+log(xcosx)]+(xsinx)^{\frac{1}{x}}[\frac{xcotx+1-log(xsinx)}{x^2}]$