# What is the derivative of sin 5x?

Feb 18, 2017

$\frac{d \sin \left(5 x\right)}{\mathrm{dx}} = \textcolor{g r e e n}{5 \cos \left(5 x\right)}$

#### Explanation:

Remember the Chain Rule for Derivatives:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{\frac{d f \left(g \left(x\right)\right)}{d x}}$=color(blue)((d f(g(x)))/(d g(x)) * (d g(x))/(dx)

If we let
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b r o w n}{f \left(x\right) = \sin \left(x\right)}$ and
$\textcolor{w h i t e}{\text{XXX}} \textcolor{m a \ge n t a}{g \left(x\right) = 5 x}$
then $f \left(g \left(x\right)\right) = \sin \left(5 x\right) \textcolor{w h i t e}{\text{XX}}$ (the expression given in the question)

It is sometimes easier to replace $g \left(x\right)$ with a simple variable.
Let's do that and let $u = g \left(x\right)$
So
$\textcolor{w h i t e}{\text{XXX} \textcolor{b r o w n}{} \frac{d f \left(g \left(x\right)\right)}{d g \left(x\right)}}$ becomes $\textcolor{b r o w n}{\frac{d f \left(u\right)}{\mathrm{du}} = \frac{d \sin \left(u\right)}{d u}}$

Hopefully, you remember the basic trigonometric derivative:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b r o w n}{\frac{d \sin \left(u\right)}{d u} = \cos \left(u\right) = \cos \left(g \left(x\right)\right) = \cos \left(5 x\right)}$

Also, since $\textcolor{m a \ge n t a}{g \left(x\right) = 5 x}$
$\textcolor{w h i t e}{\text{XXX}} \textcolor{m a \ge n t a}{\frac{d \left(g \left(x\right)\right)}{d x} = \frac{d 5 x}{d x} = 5}$

Therefore:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{\frac{d f \left(g \left(x\right)\right)}{d x}}$=color(blue)((d f(g(x)))/(d g(x)) * (d g(x))/(dx)=color(magenta)(5) * color(brown)(cos(5x))