# How do I find the derivative of a fraction?

May 1, 2018

We use quotient rule as described below to differentiate algebraic fractions or any other function written as quotient or fraction of two functions or expressions

#### Explanation:

When we are given a fraction say $f \left(x\right) = \frac{3 - 2 x - {x}^{2}}{{x}^{2} - 1}$. This comprises of two fractions - say one $g \left(x\right) = 3 - 2 x - {x}^{2}$ in numerator and the other $h \left(x\right) = {x}^{2} - 1$, in the denominator. Here we use quotient rule as described below.

Quotient rule states if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right) - \frac{\mathrm{dh}}{\mathrm{dx}} \times g \left(x\right)}{h \left(x\right)} ^ 2$

Here $g \left(x\right) = 3 - 2 x - {x}^{2}$ and hence $\frac{\mathrm{dg}}{\mathrm{dx}} = - 2 - 2 x$ and as $h \left(x\right) = {x}^{2} - 1$, we have $\frac{\mathrm{dh}}{\mathrm{dx}} = 2 x$ and hence

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\left(- 2 - 2 x\right) \times \left({x}^{2} - 1\right) - 2 x \times \left(3 - 2 x - {x}^{2}\right)}{{x}^{2} - 1} ^ 2$

= $\frac{- 2 {x}^{3} - 2 {x}^{2} + 2 x + 2 - 6 x + 4 {x}^{2} + 2 {x}^{3}}{{x}^{2} - 1} ^ 2$

= $\frac{2 {x}^{2} - 4 x + 2}{{x}^{2} - 1} ^ 2$

or $\frac{2 {\left(x - 1\right)}^{2}}{{x}^{2} - 1} ^ 2$

= $\frac{2}{x + 1} ^ 2$

Observe that $\frac{3 - 2 x - {x}^{2}}{{x}^{2} - 1} = \frac{\left(1 - x\right) \left(3 + x\right)}{\left(x + 1\right) \left(x - 1\right)} = \frac{- 3 - x}{x + 1}$ and using quotient rule

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{- \left(x + 1\right) - \left(- 3 - x\right)}{x + 1} ^ 2 = \frac{2}{x + 1} ^ 2$