# How do you find f'(0) given f(x) =( (x^5+5)/(3cosx))^2?

Dec 22, 2016

$f ' \left(0\right) = 2 \left(\frac{{0}^{5} + 5}{3 \cos 0}\right) \cdot 0 = 0$ See below.

#### Explanation:

$f ' \left(x\right) = 2 \left(\frac{{x}^{5} + 5}{3 \cos x}\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{{x}^{5} + 5}{3 \cos x}\right)$

Notice that when we apply the quotient rule to find

$\frac{d}{\mathrm{dx}} \left(\frac{{x}^{5} + 5}{3 \cos x}\right)$,

The derivatives of the numerator ($5 {x}^{4}$) and denominator $\left(- 3 \sin x\right)$ are both $0$ at $x = 0$, so lets not bother with it.

$f ' \left(0\right) = 2 \left(\frac{{0}^{5} + 5}{3 \cos 0}\right) \cdot 0 = 0$