Question

Use implicit differentiation to find dy/dx if

x^2y^3+x^3=3y^4+1

Answer 1

\[\huge x^2y^3 + x^3 = 3y^4 +1\]

\[\huge 2xy^3 + 3y^2y'x^2 + 3x^2 = 12y^3y'\]
\[\huge 2xy^3 + 3x^2 = 12y^3y' - 3y^2y'x^2\]
\[\huge 2xy^3 + 3x^2 = (12y^3 - 3y^2x^2)y'\]
now solve for y'

Answer 2

This will take a moment.

Answer 3

Trying to find a theoretical way to help you understand how implicit differentiation works.

Answer 4

\[\huge \frac {2xy^3 + 3x^2}{(12y^3 - 3y^2x^2)} = y'\]

Answer 5

I would suggest to look at partial derivatives, it is a bit more advanced, but I found that when I took them, it made implicit differentiation easy. Basically, if differentiating a function of x and y with respect to y, you will hold x as you would a constant, and then do differentation as normal. Sometimes it is helpful to do this by breaking the problem into smaller steps, and remembering the product and quotient and chain rules. Hope this helps.

Answer 6

Thanks dpainc and seanwalsh!

Answer 7

They teach product rule before they teach partial derivatives.

Just treat x and y as 2 different functions of some generic variable (*) and derive to your hearts content.

in the end, change your generic * to the required x in this case.

dx/d* becomes dx/dx = 1

dy/d* becomes dy/dx

Answer 8

\[\frac{d}{d*}xy=x\frac{dy}{d*}+\frac{dx}{d*}y\]

\[\frac{d}{dx}xy=x\frac{dy}{dx}+\frac{dx}{dx}y\]
\[\frac{d}{dx}xy=xy'+y\]