# How do you find the derivative of (t^1.7 + 8)/(t^1.4 + 6)?

May 14, 2015

Quocient rule!

Let's just remember the quocient rule by definition, here:

be $y = \frac{f \left(x\right)}{g \left(x\right)}$, then its derivative is given by:

$y ' = \frac{f ' \left(x\right) \cdot g \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{f {\left(x\right)}^{2}}$.

Now, let's just derivate your function accordingly:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1.7 {t}^{0.7} \cdot \left({t}^{1.4} + 6\right) - \left({t}^{1.7} + 8\right) \left(1.4 {t}^{0.4}\right)}{{t}^{1.7} + 8} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1.7 {t}^{2.1} + 10.2 {t}^{0.7}\right) - 1.4 {t}^{2.1} + 11.2 {t}^{0.4}}{{t}^{1.7} + 8} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{0.3 {t}^{2.1} + 10.2 {t}^{0.7} - 11.2 {t}^{0.4}}{{t}^{1.7} + 8} ^ 2$