# How do you differentiate y=(x^3-x^2-3)/(x^5+3) using the quotient rule?

Mar 30, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{4} + 15 {x}^{2} - 4 x}{{x}^{2} + 5} ^ 2$

#### Explanation:

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

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

Here $g \left(x\right) = {x}^{3} - {x}^{2} - 3$ hence $\frac{\mathrm{dg}}{\mathrm{dx}} = 3 {x}^{2} - 2 x$

and as $h \left(x\right) = {x}^{2} + 5$, $\frac{\mathrm{dh}}{\mathrm{dx}} = 2 x$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(3 {x}^{2} - 2 x\right) \times \left({x}^{2} + 5\right) - 2 x \times \left({x}^{3} - {x}^{2} - 3\right)}{{x}^{2} + 5} ^ 2$

= $\frac{3 {x}^{4} - 2 {x}^{3} + 15 {x}^{2} - 10 x - 2 {x}^{4} + 2 {x}^{3} + 6 x}{{x}^{2} + 5} ^ 2$

= $\frac{{x}^{4} + 15 {x}^{2} - 4 x}{{x}^{2} + 5} ^ 2$