# How do you differentiate f(x)=1/(x-1)^2+x^3-4/x using the sum rule?

Jan 22, 2016

$f ' \left(x\right) = 3 {x}^{2} + \frac{4}{x} ^ 2 - \frac{2}{x - 1} ^ 3$

#### Explanation:

The sum rule states that we can find the derivative of each added part individually and then combine them, like so:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[\frac{1}{x - 1} ^ 2\right] + \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] - \frac{d}{\mathrm{dx}} \left[\frac{4}{x}\right]$

Finding the derivative of the first term: we can redefine the function using negative exponents, and then differentiate using the chain rule.

$\frac{d}{\mathrm{dx}} \left[{\left(x - 1\right)}^{-} 2\right] = - 2 {\left(x - 1\right)}^{-} 3 \frac{d}{\mathrm{dx}} \left[x - 1\right]$

=-2(x-1)^-3(1)=color(red)(-2/(x-1)^3

The second term will require a simple application of the chain rule.

d/dx[x^3]=color(red)(3x^2

The third can also be rewritten with negative exponents.

d/dx[4x^-1]=-1(4x^-2)=color(red)(-4/x^2

Thus, the derivative of the function is

$f ' \left(x\right) = - \frac{2}{x - 1} ^ 3 + 3 {x}^{2} - \left(- \frac{4}{x} ^ 2\right)$

Slightly simplified/reordered:

$f ' \left(x\right) = 3 {x}^{2} + \frac{4}{x} ^ 2 - \frac{2}{x - 1} ^ 3$