Question

If $y=\sin \left(x^x\right)$ then $\frac{dy}{dx}=$

• A. $x^x\cos \left(x^x\right)(1-\log x)$
• B. $x^x\bullet \sin \left(x^x\right)(1-\log x)$
• C. $x^x\cos \left(x^x\right)(1+\log x)$
• D. $x^x\sin \left(x^x\right)(1+\log x)$

Answer 1

The correct option is C $x^x\cos \left(x^x\right)(1+\log x)$

We have,

$y=\sin x^x$

On differentiating both sides w.r.t $x$, we get

$\frac{dy}{dx}=\frac{d}{dx}\left(\sin x^x\right)$

$\frac{dy}{dx}=\cos \left(x^x\right)\frac{d}{dx}\left(x^x\right)$

$\frac{dy}{dx}=\cos \left(x^x\right)\left(x^x\right)\frac{d}{dx}\left(x\ln x\right)$

$\frac{dy}{dx}=\left(x^x\right)\cos \left(x^x\right)(\ln x\times 1+x\times \frac{1}{x})$

$\frac{dy}{dx}=\left(x^x\right)\cos \left(x^x\right)(\ln x+1)$

Hence, this is the answer.