find the derivative of the function. f(t) = 4√2 + tan t f(t) =

Explanation:

Step1: Rewrite the function

Rewrite $f(t)=\sqrt[4]{2 + \tan t}=(2+\tan t)^{\frac{1}{4}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^n$ and $u = g(t)$, then $\frac{dy}{dt}=n\cdot u^{n - 1}\cdot\frac{du}{dt}$. Here $n=\frac{1}{4}$ and $u = 2+\tan t$. First, find $\frac{du}{dt}$. Since $\frac{d}{dt}(2)=0$ and $\frac{d}{dt}(\tan t)=\sec^{2}t$, then $\frac{du}{dt}=\sec^{2}t$.

Step3: Calculate the derivative

$f'(t)=\frac{1}{4}(2 + \tan t)^{\frac{1}{4}-1}\cdot\sec^{2}t=\frac{1}{4}(2+\tan t)^{-\frac{3}{4}}\cdot\sec^{2}t=\frac{\sec^{2}t}{4(2 + \tan t)^{\frac{3}{4}}}$.

Answer:

$\frac{\sec^{2}t}{4(2+\tan t)^{\frac{3}{4}}}$