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. Thenwe find the derivative of fg by multiplying it by the derivativeof y cubed, which is 3y2. Since we are finding the derivative withrespect to x and we are dealing with the variable y, we mustmultiply this result by dy/dx =3y2. Finishing up our product, the derivative of f, that's 2x times ycubed. And then the derivative of y to the fourth is 4y cubed, andthen again, dy/dx. Derivative of x, that's 1. Now isolate all terms that have a dy/dx in them. There are twoterms with dy/dx's. Anything that doesn't have dy/dx needs to bemoved to the other side of the equation. Left versus right doesnot matter when isolating variables. For simplicity, we have two dy/dx's already on the left side. There is 3x squared y squared dxminus 4y cubed dy/dx, and that equals 1 minus 2xy cubed.The next step is to isolate dy/dx. In this case, we will factorout a dy/dx from our equation and multiply the result by 3xsquared y squared minus 4y cubed. This gives us minus 2xy cubeddivided by 3x squared y squared minus 4y cubed. To finish, wesimply divide by this term. The final answer is dy/dx equals 1minus 2xy cubed divided by 3x squared y squared minus 4y cubed.