How do you implicitly differentiate #-y= x^3y^2-x^2y^3-2xy^4 #?

Question

How do you implicitly differentiate #-y= x^3y^2-x^2y^3-2xy^4    #?

Evelyn Agard · Accepted Answer

#y'=(3x^2y^2-2xy^3-2y^4)/(3x^2y^2+8xy^3-2x^3y-1)#

From the given #-y=x^3y^2-x^2y^3-2xy^4# differentiate both sides of the equation with respect to x

#d/dx(-y)=d/dx(x^3y^2-x^2y^3-2xy^4)#

#-y'=x^3*d/dx(y^2)+y^2*d/dx(x^3)-[x^2*d/dx(y^3)+y^3*d/dx(x^2)]-2*[x*d/dx(y^4)+y^4*d/dx(x)]#

#-y'=x^3*2yy'+3x^2y^2-3x^2y^2y'-2xy^3-8xy^3y'-2y^4#

Isolate all terms with y' in one side of the equation

#-y'-2x^3yy'+3x^2y^2y'+8xy^3y'=3x^2y^2-2xy^3-2y^4#

Factor out the common monomial factor #y'#

#(-1-2x^3y+3x^2y^2+8xy^3)y'=3x^2y^2-2xy^3-2y^4#

Divide both sides of the equation by the coefficient of #y'# which is #(-1-2x^3y+3x^2y^2+8xy^3)#

#((-1-2x^3y+3x^2y^2+8xy^3)y')/((-1-2x^3y+3x^2y^2+8xy^3))=(3x^2y^2-2xy^3-2y^4)/(-1-2x^3y+3x^2y^2+8xy^3)#

#y'=(3x^2y^2-2xy^3-2y^4)/(3x^2y^2+8xy^3-2x^3y-1)#

God bless....I hope the explanation is useful.

Isabella Ackerman · Answer

undefined To implicitly differentiate the given equation $-y = x^3y^2 - x^2y^3 - 2xy^4$, you can follow these steps:

1. Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$.
2. Apply the product rule and chain rule as necessary on the right side.
3. Collect terms involving $\frac{dy}{dx}$ on one side and solve for $\frac{dy}{dx}$.

After performing these steps, the derivative $\frac{dy}{dx}$ is found to be:

$$
\frac{dy}{dx} = \frac{y^2(-3x^2 + 2xy + 2y^2)}{2x^3y - 3x^2y^2 - 8xy^3 + 1}
$$