Question

find the derivative of the function.
$f(x)=\tan ^{-1}(e^{4 x})$
$f^{prime}(x)=\square$

Explanation:

Step1: Recall the chain rule

The chain rule states that if \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). For \(y = \tan^{- 1}(u)\), the derivative is \(y^\prime=\frac{1}{1 + u^{2}}\cdot u^\prime\), where \(u = e^{4x}\).

Step2: Find the derivative of \(u = e^{4x}\)

Using the formula \((e^{ax})^\prime=ae^{ax}\), for \(u = e^{4x}\), we have \(u^\prime = 4e^{4x}\).

Step3: Apply the chain rule

Substitute \(u = e^{4x}\) and \(u^\prime=4e^{4x}\) into the formula for the derivative of \(\tan^{-1}(u)\).

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Answer:

\(\frac{4e^{4x}}{1 + e^{8x}}\)