Question

Differentiate the function with respect to $x$.
$x^x-2^{sinx}$

Answer 1

The correct answer is $∴\frac{dy}{dx}=x^x(1+log\,x)-2^{sinx}cos\,xlog\,2$
Let $y=x^x-2^{sinx}$
Also,let $x^x=u$ and $2^{sinx}=v$
$∴y=u-v$
$⇒\frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}$
$u=x^x$
Taking logarithm on both the sides,we obtain
$log\,u=xlog\,x$
Differentiating both sides with respect to $x$,we obtain
$\frac{1}{u}\frac{du}{dx}=[\frac{d}{dx}(x)\times log\,x+x\times \frac{d}{dx}(logx)]$
$⇒\frac{du}{dx}=u[1\times log\,x+x\times \frac{1}{x}]$
$⇒\frac{du}{dx}=x^x(logx+1)$
$⇒\frac{du}{dx}=x^x(1+log\,x)$
$v=2^{sinx}$
Taking logarithm on both the sides with respect to $x$,we obtain
$log\,v=sin\,x.log\,2$
Differentiating both sides with respect to $x$,we obtain
$\frac{1}{v}\frac{dv}{dx}=log2.\frac{d}{dx}(sin\,x)$
$⇒\frac{dv}{dx}=v\,log\,2\,cosx$
$⇒\frac{dv}{dx}=2^{sinx}cos\,xlog\,2$
$∴\frac{dy}{dx}=x^x(1+log\,x)-2^{sinx}cos\,xlog\,2$