# How do you differentiate f(x)= (2 x^2 + 7 x - 2)/ (x - cos x ) using the quotient rule?

Jul 14, 2017

d/(dx) [(2x^2+7x-2)/(x-cosx)] = color(blue)((4x+7)/(x-cosx) - ((2x^2 + 7x - 2)(1+sinx))/((x-cosx)^2)

#### Explanation:

We're asked to find the derivative

$\frac{d}{\mathrm{dx}} \left[\frac{2 {x}^{2} + 7 x - 2}{x - \cos x}\right]$

Use the quotient rule, which is

$\frac{d}{\mathrm{dx}} \left[\frac{u}{v}\right] = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{{v}^{2}}$

where

• $u = 2 {x}^{2} + 7 x - 2$

• $v = x - \cos x$:

$= \frac{\left(x - \cos x\right) \left(\frac{d}{\mathrm{dx}} \left[2 {x}^{2} + 7 x - 2\right]\right) - \left(2 {x}^{2} + 7 x - 2\right) \left(\frac{d}{\mathrm{dx}} \left[x - \cos x\right]\right)}{{\left(x - \cos x\right)}^{2}}$

The derivative of $2 {x}^{2} + 7 x - 2$ is $4 x + 7$ (use power rule for each term):

$= \frac{\left(x - \cos x\right) \left(4 x + 7\right) - \left(2 {x}^{2} + 7 x - 2\right) \left(\frac{d}{\mathrm{dx}} \left[x - \cos x\right]\right)}{{\left(x - \cos x\right)}^{2}}$

The derivative of $x$ is $1$ (power rule) and the derivative of $\cos x$ is $- \sin x$:

= color(blue)(((x-cosx)(4x+7)-(2x^2 + 7x - 2)(1+sinx))/((x-cosx)^2)

Which can also be written as

= color(blue)((4x+7)/(x-cosx) - ((2x^2 + 7x - 2)(1+sinx))/((x-cosx)^2)