How do you differentiate y=(x+1)/(x^3+x-2)?

May 29, 2018

Below

Explanation:

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{3} + x - 2 - 3 {x}^{3} - x - 3 {x}^{2} - 1}{{x}^{3} + x - 2} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 {x}^{3} - 3 {x}^{2} - 3}{{x}^{3} + x - 2} ^ 2$

The quotient rule is given by:

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

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

In this case, your u is $x + 1$ and your v is ${x}^{3} + x - 2$