# How do you implicitly differentiate  sqrt(3x+3y) + sqrt(3xy) = 17.5?

Nov 26, 2016

$\sqrt{3 x + 3 y} + \sqrt{3 x y} = 17.5$

$\sqrt{3 x + 3 y} + \sqrt{3 x y} = \sqrt{3} \sqrt{x + y} + \sqrt{3} \sqrt{x y}$, so

$\sqrt{3 x + 3 y} + \sqrt{3 x y} = 17.5$ is equivalent to

$\sqrt{x + y} + \sqrt{x y} = \frac{17.5}{\sqrt{3}}$

Differentiate both sides with respect to $x$, using the chain rule.

$\frac{1}{2 \sqrt{x + y}} \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \frac{1}{2 \sqrt{x y}} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

1/(2sqrt(x+y))+ 1/(2sqrt(x+y))dy/dx + y/(2sqrt(xy))+x/(2sqrt(xy)) dy/dx) = 0

((sqrt(xy)+xsqrt(x+y))/(sqrt(xy)sqrt(x+y)))dy/dx = (-sqrt(xy)-ysqrt(x+y))/(sqrt(xy)sqrt(x+y)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \sqrt{x y} - y \sqrt{x + y}}{\sqrt{x y} + x \sqrt{x + y}}$