# How do you differentiate f(x)=sqrt(cose^(4x) using the chain rule.?

$f ' \left(x\right) = \frac{- 2 \cdot {e}^{4 x} \cdot \sin {e}^{4 x}}{\sqrt{\cos {e}^{4 x}}}$

#### Explanation:

We start from the given
$f \left(x\right) = \sqrt{\cos {e}^{4 x}}$

$f ' \left(x\right) = \frac{1}{2 \sqrt{\cos {e}^{4 x}}} \cdot \frac{d}{\mathrm{dx}} \left(\cos {e}^{4 x}\right)$

$f ' \left(x\right) = \frac{1}{2 \sqrt{\cos {e}^{4 x}}} \cdot \left(- \sin {e}^{4 x}\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{4 x}\right)$

$f ' \left(x\right) = \frac{1}{2 \sqrt{\cos {e}^{4 x}}} \cdot \left(- \sin {e}^{4 x}\right) \left({e}^{4 x}\right) \frac{d}{\mathrm{dx}} \left(4 x\right)$

$f ' \left(x\right) = \frac{1}{2 \sqrt{\cos {e}^{4 x}}} \cdot \left(- \sin {e}^{4 x}\right) \left({e}^{4 x}\right) \cdot 4$

$f ' \left(x\right) = \frac{- 2 \cdot {e}^{4 x} \cdot \sin {e}^{4 x}}{\sqrt{\cos {e}^{4 x}}}$

God bless....I hope the explanation is useful.