Question

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

• A. $y\left(\log \right(\sin x)+x\cot x)$
• B. $y\left(\log \right(\sin x)-x\cot x)$
• C. $-y\left(\log \right(\sin x)-x\cot x)$
• D. $-y\left(\log \right(\sin x)+x\cot x)$

Answer 1

The correct option is A $y\left(\log \right(\sin x)+x\cot x)$

$y={\left(sinx\right)}^x$

$logy=xlogsinx$

$\frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}\left(logsinx\right)+logsin\frac{d}{dx}\left(x\right)$

$\frac{1}{y}\frac{dy}{dx}=\frac{x}{sinx}cosx+logsinx$

$\frac{1}{y}\frac{dy}{dx}=xcotx+logsinx$

$\frac{dy}{dx}=y(xcotx+logsinx)$