Let $f(x)=\sec ^{-1}\left(4^{x}\right)$. Find $f^{\prime}(x)$.
\[
f^{\prime}(x)=
\]

Step 1 ：Given the function \(f(x)=\sec ^{-1}\left(4^{x}\right)\)

Step 2 ：We need to find its derivative using the chain rule and the derivative of the inverse secant function

Step 3 ：The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function

Step 4 ：The derivative of the inverse secant function is \(\frac{1}{|x|\sqrt{x^2-1}}\)

Step 5 ：First, find the derivative of the inner function, \(4^x\). The derivative of \(a^x\) is \(a^x \ln(a)\), so the derivative of \(4^x\) is \(4^x \ln(4)\)

Step 6 ：Now, apply the chain rule: \(f^\prime(x) = \frac{1}{|4^x|\sqrt{(4^x)^2-1}} \cdot 4^x \ln(4)\)

Step 7 ：Simplify the expression: \(f^\prime(x) = \frac{4^x \ln(4)}{|4^x|\sqrt{(4^x)^2-1}}\)

Step 8 ：Since \(4^x\) is always positive, we can remove the absolute value: \(f^\prime(x) = \frac{4^x \ln(4)}{4^x\sqrt{(4^x)^2-1}}\)

Step 9 ：Simplify further: \(f^\prime(x) = \frac{\ln(4)}{\sqrt{(4^x)^2-1}}\)

Step 10 ：So, the derivative of the function \(f(x)=\sec ^{-1}\left(4^{x}\right)\) is \(\boxed{\frac{\ln(4)}{\sqrt{(4^x)^2-1}}}\)