How do you differentiate  y = (5x^2) / (4x-1)?

Jun 9, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{10 x \left(2 x - 1\right)}{4 x - 1} ^ 2$

Just use the quotient rule and simplify.

Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(4 x - 1\right) \cdot \frac{d}{\mathrm{dx}} \left[5 {x}^{2}\right] - 5 {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left[4 x - 1\right]}{4 x - 1} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(4 x - 1\right) \cdot 10 x - 5 {x}^{2} \cdot 4}{16 {x}^{2} + 1 - 8 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(40 {x}^{2} - 10 x\right) - 20 {x}^{2}}{16 {x}^{2} + 1 - 8 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{10 x \left(2 x - 1\right)}{4 x - 1} ^ 2$