find the derivative of the function.\n$f(x)=\\tan ^{-1}(e^{4 x})$\n$f^{prime}(x)=\\square$
find the derivative of the function.\n$f(x)=\\tan ^{-1}(e^{4 x})$\n$f^{prime}(x)=\\square$
Answer
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)). [ \begin{align*} f^\prime(x)&=\frac{1}{1+(e^{4x})^{2}}\cdot4e^{4x}\ &=\frac{4e^{4x}}{1 + e^{8x}} \end{align*} ]
Answer:
(\frac{4e^{4x}}{1 + e^{8x}})