Question

The derivative of $x^{{\sin }^{-1}x}$ with respect to ${\sin }^{-1}x$ is

• A. $x^{{\sin }^{-1}x}[\ln x+{\sin }^{-1}x\frac{\sqrt{(1-x^2)}}{x}]$
• B. $-x^{{\sin }^{-1}x}[\ln x+{\sin }^{-1}x\frac{\sqrt{(1-x^2)}}{x}]$
• C. $x^{{\sin }^{-1}x}[\ln x+{\sin }^{-1}x\frac{\sqrt{(1+x^2)}}{x}]$
• D. $-x^{{\sin }^{-1}x}[\ln x+{\sin }^{-1}x\frac{\sqrt{(1+x^2)}}{x}]$

Answer 1

The correct option is A $x^{{\sin }^{-1}x}[\ln x+{\sin }^{-1}x\frac{\sqrt{(1-x^2)}}{x}]$

Let $y=x^{{\sin }^{-1}x}$ and $z={\sin }^{-1}x$
Taking Logarithm on both sides,
$\Rightarrow \ln y=\left({\sin }^{-1}x\right)\ln x\Rightarrow \ln y=z\ln \sin z(\because \sin z=x)$
Differentiating $y$ w.r.t. $z$, we get
$\Rightarrow \frac{1}{y}\frac{dy}{dz}=\ln \sin z+z\cot z\therefore \frac{dy}{dz}=x^{{\sin }^{-1}x}[\ln x+{\sin }^{-1}x\frac{\sqrt{(1-x^2)}}{x}](\because \cot z=\frac{\sqrt{1-x^2}}{x})$