Question

Derivative of $(\sin x{)}^x+{\sin }^{-1}\sqrt{x}$ with respect to $x$ is

• A. $(x\cot x+\log \sin x)+\frac{1}{2\sqrt{x-x^2}}$
• B. $(x\cot x+\log \sin x)+\frac{1}{\sqrt{x-x^2}}$
• C. $\left(\sin x{)}^x\right(x\cot x+\log x)+\frac{1}{\sqrt{x-x^2}}$
• D. $\left(\sin x{)}^x\right(x\cot x+\log \sin x)+\frac{1}{2\sqrt{x-x^2}}$

Answer 1

The correct option is D $\left(\sin x{)}^x\right(x\cot x+\log \sin x)+\frac{1}{2\sqrt{x-x^2}}$

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

$\Rightarrow \log y=x\log \left(\sin x\right)$

Now differentiating both sides with respect to $x$

$\frac{1}{y}\frac{dy}{dx}=x\cot x+\log \sin x$

or, $\frac{dy}{dx}=\left(\sin x{)}^x\right\{x\cot x+\log \sin x\}$........(1).

Again let $z=(\sin x{)}^x+{\sin }^{-1}\sqrt{x}$

Now differentiating both sides with respect to $x$.

$\frac{dz}{dx}=\frac{dy}{dx}+\frac{1}{\sqrt{1-x}}.\frac{1}{2\sqrt{x}}$

$\frac{dz}{dx}=\left(\sin x{)}^x\right\{x\cot x+\log \sin x\}+\frac{1}{2\sqrt{x-x^2}}$ [Using (1)]