# What is the derivative of f(t)=cos^2(3t+5)?

#### Answer:

$f ' \left(t\right) = - 6 \cdot \sin \left(3 t + 5\right) \cdot \cos \left(3 t + 5\right)$

#### Explanation:

${\cos}^{2} \left(3 t + 5\right)$
$= \cos \left(3 t + 5\right) \cdot \cos \left(3 t + 5\right)$

Use product rule:
$= \frac{d}{\mathrm{dx}} \cos \left(3 t + 5\right) \cdot \cos \left(3 t + 5\right) + \frac{d}{\mathrm{dx}} \cos \left(3 t + 5\right) \cdot \cos \left(3 t + 5\right)$

Use the chain rule to differentiate $\cos \left(3 t + 5\right)$
$= - \sin \left(3 t + 5\right) \cdot 3 \cdot \cos \left(3 t + 5\right) - \sin \left(3 t + 5\right) \cdot 3 \cdot \cos \left(3 t + 5\right)$
$= - 3 \cdot \sin \left(3 t + 5\right) \cdot \cos \left(3 t + 5\right) - 3 \cdot \sin \left(3 t + 5\right) \cdot \cos \left(3 t + 5\right)$

Simplify
$= - 6 \cdot \sin \left(3 t + 5\right) \cos \left(3 t + 5\right)$