Question

What is the slope of the tangent to the curve$\mathrm{y}={\sin }^{-1}\left({\sin }^2\mathrm{x}\right)$ at $\mathrm{x}=0$?

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Slope of the tangent:

The slope of a tangent to the curve can be determined by finding the derivative.

$\mathrm{y}={\sin }^{-1}\left({\sin }^2\left(\mathrm{x}\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}·{\sin }^{-1}\left({\sin }^2\left(\mathrm{x}\right)\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{1}{\sqrt{1-{\sin }^4\left(\mathrm{x}\right)}}\right]\left(2\sin \left(\mathrm{x}\right)·\cos \left(\mathrm{x}\right)\right)$

Put the value of $x=0$.

$\frac{d\mathrm{y}}{d\mathrm{x}}=\left[\frac{1}{\sqrt{1-{\sin }^4\left(0\right)}}\right]\left(2\sin \left(0\right)·\cos \left(0\right)\right)$

$\Rightarrow \frac{d\mathrm{y}}{d\mathrm{x}}=1·0$

$\Rightarrow \frac{d\mathrm{y}}{d\mathrm{x}}=0$

So, the slope of the tangent to the curve$\mathrm{y}={\sin }^{-1}\left({\sin }^2\mathrm{x}\right)$ at $\mathrm{x}=0$ is $0$.

Hence, the correct answer is option (A) .