Question

Encontre dy/dx derivando implicitamente: x^(4 ) ( x+y)= y^2 (3x-y )

Answer 1

$x^4(x+y)= y^2(3x-y)$
$x^5+x^4y=y^23x-y^3$

Derivando:

$\frac{d(x^5)}{dx} + \frac{d(x^4y)}{dx}= \frac{d(y^23x)}{dx}- \frac{d(y^3)}{dx} \\ \\ 5x^4+(4x^3y+x^4.y' )=(2y.y'.3x+y^2.3)-3y^2.y' \\ \\ 5x^4+4x^3y+x^4.y' =6xy.y'+3y^2-3y^2.y' \\ \\ x^4y'-6xy.y' + 3y^2.y' = 3y^2-5x^4-4x^3y \\ \\ y'(x^4-6xy+ 3y^2) = 3y^2-5x^4-4x^3y \\ \\ \boxed{\boxed{y' = \frac{ 3y^2-5x^4-4x^3y}{x^4-6xy+ 3y^2} }}$