Evaluate the derivative of the function.
​f(x)equalssine Superscript negative 1 Baseline left parenthesis 9 x Superscript 4 Baseline right parenthesis

To evaluate the derivative of the function \( f(x) = \sin^{-1}(9x^4) \), we'll use the chain rule. The derivative of the inverse sine function, \( \sin^{-1}(u) \), with respect to \( u \), is:

\[
\frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}}
\]

Here, \( u = 9x^4 \). We also need to differentiate \( u = 9x^4 \) with respect to \( x \), which is:

\[
\frac{d}{dx}(9x^4) = 36x^3
\]

Now, applying the chain rule:

\[
f'(x) = \frac{1}{\sqrt{1 - (9x^4)^2}} \cdot 36x^3
\]

Simplifying the expression inside the square root:

\[
f'(x) = \frac{36x^3}{\sqrt{1 - 81x^8}}
\]

Thus, the derivative of the function is:

\[
f'(x) = \frac{36x^3}{\sqrt{1 - 81x^8}}
\]

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. What is the second derivative of \( f(x) = \sin^{-1}(9x^4) \)?
2. How would you find the derivative of \( \cos^{-1}(9x^4) \)?
3. Can you compute the derivative of \( f(x) = \sin^{-1}(x^n) \) for general \( n \)?
4. How does the derivative of \( \tan^{-1}(9x^4) \) differ from that of \( \sin^{-1}(9x^4) \)?
5. What is the domain of \( f(x) = \sin^{-1}(9x^4) \) for the derivative to be real?

**Tip:** Remember that the inverse sine function only works for values in the range \( [-1, 1] \), so always check the domain when dealing with derivatives of inverse trigonometric functions.

Evaluate the derivative of the function f(x) = sin^(-1)(9x^4).

Learn how to evaluate the derivative of the function f(x) = sin^(-1)(9x^4) using the chain rule and inverse trigonometric function properties. This step-by-step guide explains the key concepts and formulas required for the solution, suitable for advanced high school or college calculus students.

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