How to Apply the Derivative Product Rule

Learn to apply the derivative product rule in calculus. The derivative of the sum of two functions is the sum of the derivatives. However, all things are not equal. The derivative of the product of two functions is "not" the product of two derivatives. But fret not. There is a simple formula for the product of two derivatives that you can commit to memory to make those calculations a cinch.

Instructions

1. Look at the following two examples and you see how you get two different answers: d/dt(t x t) = d/dt(t^2) = 2t, but d/dt(t) x d/dt(t) = 1 x 1 = 1
   This is a sure give away that there must be a better way to find derivatives.

2. Memorize the derivative product rule: If u and v are differentiable at t, then the product uv is also differentiable, and d/dt (uv) = (u) dv/dt + (v) du/dt. This is the formula to use to find derivatives of products.

3. Find the derivative for y = (x^3 + 3)(x^2 + 1) now in Step 4 and Step 5. Both of these methods use the derivative product rule noted in Step 2. They are a little different, however, and alternate methods in a way. Both ways work.

4. Use the derivative product rule first: u = (x^3 + 3) and v = (x^2 + 1). Now we work it out: d/dx[x^3 + 3)(x^2 + 1)] = (x^3 + 3)(2x) + (x^2 + 1)(3x^2) = 2x^4 + 6x + 3x^4 + 3x^2 = 5x^4 + 3x^2 + 6x

5. Try it this way now. Take the original equation and multiply it out first. y = (x^3 + 3)(x^2 + 1) = x^5 + x^3 + 3x^2 + 3. Now apply the derivative product rule: dy/dx = 5x^4 + 3x^2 + 6x. You get the same answer as in Step 5, but a little quicker.