Question

Calcule a derivada implicitamente.
(X^2 + Y^2)^3/2 = 10xy

Answer 1

$\displaystyle \bigg( x^2 + y^2 \bigg)^{ \displaystyle \frac{3}{2}} = 10xy \\ \\ \\ \frac{3}{2} \cdot \bigg( x^2 + y^2 \bigg)^{ \displaystyle \frac{3}{2} - 1} \cdot \, \, (x^2+y^2)' = 10x' \cdot y + 10x \cdot y' \\ \\ \\ \frac{3}{2} \cdot \bigg( x^2 + y^2 \bigg)^{ \displaystyle \frac{1}{2}} \cdot \, \, (2x+2y \, \frac{dy}{dx}) = 10y+10x \, \frac{dy}{dx} \\ \\ \\ \frac{3}{2}\sqrt{x^2+y^2} \cdot (2x+2y\, \frac{dy}{dx}) = 10y+10x \, \frac{dy}{dx}$

$\displaystyle 3x\sqrt{x^2+y^2} + 3y\sqrt{x^2+y^2} \, \frac{dy}{dx} = 10y+10x \, \frac{dy}{dx} \\ \\ \\ 3y\sqrt{x^2+y^2} \, \frac{dy}{dx} - 10x \, \frac{dy}{dx} = 10y-3x\sqrt{x^2+y^2} \\ \\ \\ \frac{dy}{dx} \cdot (3y\sqrt{x^2+y^2}-10x) = 10y-3x\sqrt{x^2+y^2} \\ \\ \\ \frac{dy}{dx} = \frac{ 10y-3x\sqrt{x^2+y^2}}{3y\sqrt{x^2+y^2}-10x}$