Question

Differentiate

$x^x-2\sin x$

Answer 1

Let $y=x^x-2\sin x$

Again Let $x^x=u$

Then $y=u-2\sin x$

$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}-2\cos x$ ---(1)

Now

$u=x^x$

Taking $\log$ on both sides

$\log u=x\log x$

Differentiation w.r.t, $x$

$\frac{1}{u}\frac{du}{dx}=1.\log x+x\frac{1}{x}$

$\Rightarrow \frac{du}{dx}=u(1+\log x)$

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

Putting value of $\frac{du}{dx}$ in (1) we get

$\frac{dy}{dx}=x^x(1+\log x)-2\cos x$