# How do you find the derivative of sin(cos(6x))?

Jan 18, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 6 \cos \left(\cos \left(6 x\right)\right) \sin \left(6 x\right)$

#### Explanation:

We use the chain rule a bunch.

$y = \sin \left(u\right)$, $u = \cos \left(v\right)$, $v = 6 x$.

The chain rule says:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \setminus \cdot \frac{\mathrm{du}}{\mathrm{dv}} \setminus \cdot \frac{\mathrm{dv}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{du}} = \cos \left(u\right) = \cos \left(\cos \left(v\right)\right) = \cos \left(\cos \left(6 x\right)\right)$
$\frac{\mathrm{du}}{\mathrm{dv}} = - \sin \left(v\right) = - \sin \left(6 x\right)$
$\frac{\mathrm{dv}}{\mathrm{dx}} = 6$

So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(\cos \left(6 x\right)\right) \setminus \cdot \left(- \sin \left(6 x\right)\right) \setminus \cdot 6$

cleaning up:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 6 \cos \left(\cos \left(6 x\right)\right) \sin \left(6 x\right)$

Jan 18, 2018

$- 6 \cos \left(\cos \left(6 x\right)\right) \sin \left(6 x\right)$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(h(x)))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(h \left(x\right)\right)\right) \times g ' \left(h \left(x\right)\right) \times g ' \left(x\right)$

rArrd/dx(sin(cos(6x))

$= \cos \left(\cos \left(6 x\right)\right) \times \frac{d}{\mathrm{dx}} \left(\cos \left(6 x\right)\right) \times \frac{d}{\mathrm{dx}} \left(6 x\right)$

$= \cos \left(\cos \left(6 x\right)\right) \left(- \sin \left(6 x\right)\right) \left(6\right)$

$= - 6 \cos \left(\cos \left(6 x\right)\right) \sin \left(6 x\right)$