If $f(x)=\sin ^{-1}(x)$, then what is the value of $f^{\prime}\left(\frac{\sqrt{7}}{4}\right)$ in simplest form?
Answer
\(\boxed{\frac{4}{3}}\) is the value of $f'\left(\frac{\sqrt{7}}{4}\right)$ in simplest form.
Step 1 ：Given the function $f(x) = \sin^{-1}(x)$, we need to find the derivative $f'(x)$ and then evaluate it at $x = \frac{\sqrt{7}}{4}$.
Step 2 ：Using the chain rule, we find the derivative of $f(x)$: $f'(x) = \frac{1}{\sqrt{1 - x^2}}$.
Step 3 ：Substitute $x = \frac{\sqrt{7}}{4}$ into the derivative: $f'\left(\frac{\sqrt{7}}{4}\right) = \frac{1}{\sqrt{1 - \left(\frac{\sqrt{7}}{4}\right)^2}}$.
Step 4 ：Simplify the expression: $f'\left(\frac{\sqrt{7}}{4}\right) = \frac{4}{3}$.
Step 5 ：\(\boxed{\frac{4}{3}}\) is the value of $f'\left(\frac{\sqrt{7}}{4}\right)$ in simplest form.
