# How do you find dy/dx given x = t^2 - 2t and y = t^4 - 4t?

Jun 20, 2016

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

#### Explanation:

For parametric form of equation, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$.

Here as $x = {t}^{2} - 2 t$, $\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t - 2 = 2 \left(t - 1\right)$

and as $y = {t}^{4} - 4 t$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 4 {t}^{3} - 4 = 4 \left({t}^{3} - 1\right) = 4 \left(t - 1\right) \left({t}^{2} + t + 1\right)$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \left(t - 1\right) \left({t}^{2} + t + 1\right)}{2 \left(t - 1\right)} = 2 \left({t}^{2} + t + 1\right)$