Question

if f(x) = sin^(-1)x, then f((√3)/2) =
a π/6
b π/3
c 4/7
d 2

Explanation:

Step1: Recall derivative formula

The derivative of $y = \sin^{-1}x$ is $y'=\frac{1}{\sqrt{1 - x^{2}}}$.

Step2: Substitute $x=\frac{\sqrt{3}}{2}$

Substitute $x = \frac{\sqrt{3}}{2}$ into $y'=\frac{1}{\sqrt{1 - x^{2}}}$, we get $y'=\frac{1}{\sqrt{1-(\frac{\sqrt{3}}{2})^{2}}}$.
First, calculate $(\frac{\sqrt{3}}{2})^{2}=\frac{3}{4}$. Then $1 - (\frac{\sqrt{3}}{2})^{2}=1-\frac{3}{4}=\frac{1}{4}$.
So $\sqrt{1 - (\frac{\sqrt{3}}{2})^{2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$.
Then $\frac{1}{\sqrt{1 - (\frac{\sqrt{3}}{2})^{2}}}=\frac{1}{\frac{1}{2}} = 2$.

Answer:

D. 2