if f(x) = sin^(-1)x, then f((√3)/2) =\na π/6\nb π/3\nc 4/7\nd 2
if f(x) = sin^(-1)x, then f((√3)/2) =\na π/6\nb π/3\nc 4/7\nd 2
Answer
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