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

Dec 7, 2015

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

Explanation:

Your chain can be defined as follows:

$f \left(x\right) = \textcolor{g r e e n}{\cos} \textcolor{b l u e}{{e}^{\textcolor{red}{4 x}}}$

$\textcolor{w h i t e}{\times x} = \cos {e}^{w \left(x\right)} \textcolor{w h i t e}{\times x} \text{ where } w \left(x\right) = \textcolor{red}{4 x}$

$\textcolor{w h i t e}{\times x} = \cos v \left(w\right) \textcolor{w h i t e}{\times \xi i i} \text{ where } v \left(w\right) = \textcolor{b l u e}{{e}^{w}}$

$\textcolor{w h i t e}{\times x} = u \left(v\right) \textcolor{w h i t e}{\times \times \times i} \text{ where } u \left(v\right) = \textcolor{g r e e n}{\cos \left(v\right)}$

Thus, to compute the derivative, you need to build the derivatives of $w \left(x\right)$, $v \left(w\right)$ and $u \left(v\right)$:

$w \left(x\right) = 4 x \textcolor{w h i t e}{\times x} \implies \textcolor{w h i t e}{\times} w ' \left(x\right) = 4$

$v \left(w\right) = {e}^{w} \textcolor{w h i t e}{\times x} \implies \textcolor{w h i t e}{\times} v ' \left(w\right) = {e}^{w} = {e}^{4 x}$

$u \left(v\right) = \cos v \textcolor{w h i t e}{\times} \implies \textcolor{w h i t e}{\times} u ' \left(v\right) = - \sin v = - \sin \left({e}^{w}\right) = - \sin \left({e}^{4 x}\right)$

Now, the only thing left to do is to multiply the three derivatives!

$f ' \left(x\right) = u ' \left(v\right) \cdot v ' \left(w\right) \cdot w ' \left(x\right)$
$\textcolor{w h i t e}{\times \times} = - \sin \left({e}^{4 x}\right) \cdot {e}^{4 x} \cdot 4$
$\textcolor{w h i t e}{\times \times} = - 4 \sin \left({e}^{4 x}\right) \cdot {e}^{4 x}$