Implicit Differentiation Example: Deriving dy/dx

University:
Rice University
Course:
MATH 212 | Multivariable Calculus
Academic year:

2023
Views:

322

Pages:

2
Author:

KlugInstant

Implicit Differentiation. Example Problem 1 x squared y cubed minus y to the fourth equal x. The equation above is defined implicitly. Don`t solve for y here; that would be pointless. So we'll take the derivative implicitly. As a first step, we calculate x squared times y cubed, or fg. Then we find the derivative of fg by multiplying it by the derivative of y cubed, which is 3y2. Since we are finding the derivative with respect to x and we are dealing with the variable y, we must multiply this result by dy/dx =3y2. Finishing up our product, the derivative of f, that's 2x times y cubed. And then the derivative of y to the fourth is 4y cubed, and then again, dy/dx. Derivative of x, that's 1. Now isolate all terms that have a dy/dx in them. There are two terms with dy/dx's. Anything that doesn't have dy/dx needs to be moved to the other side of the equation. Left versus right does not matter when isolating variables. For simplicity, we have two dy/dx's already on the left side. There is 3x squared y squared dx minus 4y cubed dy/dx, and that equals 1 minus 2xy cubed. The next step is to isolate dy/dx. In this case, we will factor out a dy/dx from our equation and multiply the result by 3x squared y squared minus 4y cubed. This gives us minus 2xy cubed divided by 3x squared y squared minus 4y cubed. To finish, we simply divide by this term. The final answer is dy/dx equals 1 minus 2xy cubed divided by 3x squared y squared minus 4y cubed.