Question

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find the derivative of the function.
$f(t) = \sqrt8{3 + \tan t}$
$f(t) = $
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Explanation:

Step1: Rewrite the root as exponent

$f(t)=(3+\tan t)^{1/8}$

Step2: Apply chain rule

$f'(t)=\frac{1}{8}(3+\tan t)^{-7/8} \cdot \frac{d}{dt}(3+\tan t)$

Step3: Differentiate inner function

$\frac{d}{dt}(3+\tan t)=\sec^2 t$

Step4: Simplify the expression

$f'(t)=\frac{\sec^2 t}{8(3+\tan t)^{7/8}}$

Answer:

$\frac{\sec^2 t}{8\sqrt[8]{(3+\tan t)^7}}$