Question

Find the derivative of $x^x-2^{\sin x}$ w.r.t. x

Answer 1

To find $\frac{d}{dx}(x^x-2^{\sin x})$

Let $y=x^x$

Taking $\log$ both sides we get

$\log y=x\log x$

Differentiating both sides $w.r.tx$

$\frac{1}{y}\frac{dy}{dx}=x.\frac{d}{dx}\log x+\log x\frac{d}{dx}x$

$\frac{1}{y}\frac{dy}{dx}=x.\frac{1}{x}+\log x1$

$\frac{dy}{dx}=y(1+\log x)$

$\frac{dy}{dx}=x^x(1+\log x)$

Let $z=2^{\sin x}$

Taking $\log$ both sides, we get

$\log z=\sin x\log 2$

Differentiating both sides $w.r.tx$

$\frac{1}{z}\frac{dz}{dx}=\cos x.\log 2$

$\frac{dz}{dx}=z\left(\log 2.\cos x\right)$

$\frac{dz}{dx}=2^{\sin x}\left(\log 23.\cos x\right)$

$\therefore \frac{d}{dx}(x^x-2^{\sin x})=x^x(1+\log x)-\log {2.2}^{\sin x}\cos x$