Question

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

• A. $\left(\sin x{)}^x\right(\ln \left(\sin x\right)+x\cot x)$
• B. $\left(\ln \right(\sin x)+x\cot x)$
• C. $\left(\sin x{)}^x\right(\ln \left(\sin x\right)+x\tan x)$
• D. $\left(\sin x{)}^x\right(\ln \left(\sin x\right)-\cot x)$

Answer 1

The correct option is A $\left(\sin x{)}^x\right(\ln \left(\sin x\right)+x\cot x)$

given $y=(\sin x{)}^x$

applying $\ln$ on both sides we get

$\ln y=x\ln \left(\sin x\right)$

differentiating both sides wrt $x$

$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\left(x\ln \right(\sin x\left)\right)$

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

$\frac{dy}{dx}=y\left(\ln \right(\sin x)+x\cot x)$

$\therefore \frac{dy}{dx}=(\sin x{)}^x\left(x\cot x\right)+(\sin x{)}^x\ln \left(\sin x\right)$