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

Question

Aurora Alvarado · Accepted Answer

Differentiate both sides of the equation with respect to #x#. Then isolate #dy/dx#.

Differentiate both sides of the equation with respect to #x#. By doing this, a #dy/dx# should "appear" from the #y# variable in the equation. Isolating the #dy/dx# would give you the derivative implicitly.

Differentiating both sides with respect to #x#:

#D_x[-x^(2)y^(2)-3y^(3)+2]=D_x[5x^(3)]#

This will give you:

#[(-2x)(y^(2))+(2y*dy/dx)(-x^(2))]-9y^(2)*dy/dx+0=15x^(2)#

#-2xy^(2)-2yx^(2)*dy/dx-9y^(2)*dy/dx=15x^(2)#

Put all terms with #dy/dx# to the left and shove the other terms to the right.

#-2yx^(2)*dy/dx-9y^(2)*dy/dx=15x^(2)+2xy^(2)#

Factor #dy/dx# out.

#dy/dx*(-2yx^(2)-9y^(2))= 15x^(2)+2xy^(2)#

Divide everything by #(-2yx^(2)-9y^(2))#. You will get the derivative.

#dy/dx = (15x^(2)+2xy^(2))/(-2yx^(2)-9y^(2))#

William Abdalla · Answer

undefined To find $\frac{dy}{dx}$ for the equation $-x^2y^2 - 3y^3 + 2 = 5x^3$, differentiate both sides of the equation with respect to $x$ implicitly. This yields:

$\frac{d}{dx}(-x^2y^2) + \frac{d}{dx}(-3y^3) + \frac{d}{dx}(2) = \frac{d}{dx}(5x^3)$

Simplify the derivatives:

$-2xy^2 \frac{dy}{dx} - 9y^2 \frac{dy}{dx} = 15x^2$

Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx}(-2xy^2 - 9y^2) = 15x^2$

Divide by $-2xy^2 - 9y^2$:

$\frac{dy}{dx} = \frac{15x^2}{-2xy^2 - 9y^2}$