# How do you find the derivative of [cos(2x^4) - 1]/x^7?

Mar 21, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{8}{x} ^ 4 \cos {x}^{4} + \frac{14}{x} ^ 8 {\sin}^{2} {x}^{4}$

#### Explanation:

Let
$y = \frac{\cos \left(2 {x}^{4}\right) - 1}{{x}^{7}}$

Let
$u = \cos \left(2 {x}^{4}\right) - 1$
$v = {x}^{7}$

$y = \frac{u}{v}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

$u = \cos \left(2 {x}^{4}\right) - 1$
Let
$t = {x}^{4}$

$u = \cos 2 t - 1$
wkt
$1 - \cos 2 t = 2 {\sin}^{2} t$
$u = - \left(1 - \cos 2 t\right)$
$u = - 2 {\sin}^{2} t$
By chain rule
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$

$u = - 2 {\sin}^{2} t$
${\sin}^{2} t = {\left(\sin t\right)}^{2}$
Let
$p = \sin t$

${\left(\sin t\right)}^{2} = {p}^{2}$

$u = - 2 p$
By chain rule
$\frac{\mathrm{du}}{\mathrm{dt}} = - 2 \frac{\mathrm{dp}}{\mathrm{dt}}$

$p = \sin t$

$\frac{\mathrm{dp}}{\mathrm{dt}} = \cos t$

$\frac{\mathrm{du}}{\mathrm{dt}} = - 2 \cos t$

$- 2 \cos t = - 2 \cos {x}^{4}$

$\frac{\mathrm{du}}{\mathrm{dt}} = - 2 \cos {x}^{4}$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$

$t = {x}^{4}$

$\frac{\mathrm{dt}}{\mathrm{dx}} = 4 {x}^{3}$

$\frac{\mathrm{du}}{\mathrm{dx}} = - 2 \cos {x}^{4} \left(4 {x}^{3}\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = - 8 {x}^{3} \cos {x}^{4}$

$v = {x}^{7}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = 7 {x}^{6}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

$u = - 2 {\sin}^{2} t$
$u = - 2 {\sin}^{2} {x}^{4}$
$v = {x}^{7}$
$\frac{\mathrm{du}}{\mathrm{dx}} = - 8 {x}^{3} \cos {x}^{4}$
$\frac{\mathrm{dv}}{\mathrm{dx}} = 7 {x}^{6}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{7} \left(- 8 {x}^{3} \cos {x}^{4}\right) - \left(- 2 {\sin}^{2} {x}^{4}\right) \left(7 {x}^{6}\right)}{{x}^{7}} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 8 {x}^{10} \cos {x}^{4} + 14 {x}^{6} {\sin}^{2} {x}^{4}}{x} ^ 14$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{8}{x} ^ 4 \cos {x}^{4} + \frac{14}{x} ^ 8 {\sin}^{2} {x}^{4}$