Question

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

• A. 0
• B. 1
• C. 2
• D. None of the above

Answer 1

The correct option is A 0

Slope of tangent to a curve f(x), at a point $(x_o,y_o)$ is given by $f^{\prime }\left(x\right)$,

where $f^{\prime }\left(x\right)$ is derivative of f(x) at $x=x_o$

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

$x_o=0$

$f^{\prime }\left(x\right)=\frac{df\left(x\right)}{dx}=\frac{df\left(x\right)}{d{\sin }^{2}x}\times \frac{d{\sin }^{2}x}{d\sin x}\times \frac{d\sin x}{dx}$

$f^{\prime }\left(x\right)=\frac{1}{\sqrt{1-{\sin }^{2}x}}\times 2\sin x\times \cos x$

$f^{\prime }\left(0\right)=0$

Slope of tangent to the curve is 0