# How do you find (dy)/(dx) given (2x-3)^2+(4y-5)^2=10?

Nov 6, 2016

The derivative is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 - 2 x}{8 y - 10}$.

#### Explanation:

Start by expanding the parentheses.

$4 {x}^{2} - 12 x + 9 + 16 {y}^{2} - 40 y + 25 = 10$

$\frac{d}{\mathrm{dx}} \left(4 {x}^{2} - 12 x + 9 + 16 {y}^{2} - 40 y + 25\right) = \frac{d}{\mathrm{dx}} \left(10\right)$

$\frac{d}{\mathrm{dx}} \left(4 {x}^{2}\right) + \frac{d}{\mathrm{dx}} \left(- 12 x\right) + \frac{d}{\mathrm{dx}} \left(9\right) + \frac{d}{\mathrm{dx}} \left(16 {y}^{2}\right) + \frac{d}{\mathrm{dx}} \left(- 40 y\right) + \frac{d}{\mathrm{dx}} \left(25\right) = \frac{d}{\mathrm{dx}} \left(10\right)$

$8 x - 12 + 0 + 32 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 40 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 0 = 0$

$32 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 40 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 12 - 8 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(32 y - 40\right) = 12 - 8 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{12 - 8 x}{32 y - 40}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \left(3 - 2 x\right)}{4 \left(8 y - 10\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 - 2 x}{8 y - 10}$

Hopefully this helps!