# How do you find the derivative for y= cos(x)/x^8?

Jul 30, 2015

You can use the quotient rule or the product rule.

#### Explanation:

You actually have two ways of approaching this derivative.

First, notice that your function $y$ can be written as a quotient of two other functions

$y = f \frac{x}{g} \left(x\right)$. where

$\left\{\begin{matrix}f \left(x\right) = \cos x \\ g \left(x\right) = {x}^{8}\end{matrix}\right.$

Likewsie, you can write $y$ as a product of two functions

$y = f \left(x\right) \cdot h \left(x\right)$, where

$\left\{\begin{matrix}f \left(x\right) = \cos x \\ h \left(x\right) = {x}^{- 8}\end{matrix}\right.$

You can thus differentiate this function by using

The quotient rule allows you to calculate the derivative of a function expressed as a quotient of two other functions by using the

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \frac{{f}^{'} \left(x\right) \cdot g \left(x\right) - f \left(x\right) \cdot {g}^{'} \left(x\right)}{g \left(x\right)} ^ 2}$, where $g \left(x\right) \ne 0$

In your case, the derivative of $y$ will be

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left(\cos x\right)\right] \cdot {x}^{8} - \cos x \cdot \frac{d}{\mathrm{dx}} \left({x}^{8}\right)}{{x}^{8}} ^ 2$

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{- \sin x \cdot {x}^{8} - \cos x \cdot 8 {x}^{7}}{x} ^ 16$

$\frac{d}{\mathrm{dx}} \left(y\right) = - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{7}}}} \left(x \cdot \sin x + 8 \cos x\right)}{x} ^ \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{16}}} \textcolor{b l u e}{9}\right)$

$\frac{d}{\mathrm{dx}} \left(y\right) = \textcolor{g r e e n}{- \frac{x \cdot \sin x + 8 \cdot \cos x}{x} ^ 9}$

This time ,the derivative of $y$ can be determined using the formula

color(blue)(d/dx(y) = f^'(x) * h(x) + f(x) * h^'(x)

This will get you

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(\cos x\right)\right] \cdot {x}^{- 8} + \cos x \cdot \frac{d}{\mathrm{dx}} \left({x}^{- 8}\right)$

$\frac{d}{\mathrm{dx}} \left(y\right) = - \sin x \cdot {x}^{- 8} + \cos x \cdot \left(- 8\right) \cdot {x}^{- 9}$

$\frac{d}{\mathrm{dx}} \left(y\right) = - \left(\sin \frac{x}{x} ^ 8 + \frac{8 \cos x}{x} ^ \left(9\right)\right)$

This is equivalent to

$\frac{d}{\mathrm{dx}} \left(y\right) = - \left(\frac{x \cdot \sin x}{x} ^ 9 + \frac{8 \cdot \cos x}{x} ^ 9\right) = \textcolor{g r e e n}{- \frac{x \cdot \sin x + 8 \cdot \cos x}{x} ^ 9}$