University:
Rice University
Course:
MATH 102 | Single Variable Calculus II
Academic year:

2022
Views:

7

Pages:

2
Author:

HippoGuitarist

Product Rule for Derivatives. Example Problem The function h of x equals 2x squared times f(x), where certain predefined values of f are given for x equaling 3. The question tells us to find h prime of 3, and the word "product" in the label tips us off to the fact that we must multiply two variables together. Previously, if I took the derivative of 3x to the fifth and used the product rule, technically I was multiplying two things together. But I didn't need to do that. To find the derivative of a product, such as 3(x - 1), you must use the product rule. This is because the derivative of a constant, such as 3, is 0, while the derivative of two variables multiplied together is another variable. Recall that if I take the derivative of the product of some function f, and some other function g, then I get something called f g prime plus f prime g. So arbitrarily, I'm going to call one of these functions f. I typically go by the initials. I'll call this one f. And I'll call this one g. Although it's not that big of a deal to me, I'll still get over the fact that I'm calling f of x g instead of f. If you find it difficult to follow the formula, simply memorize the steps and write down the formula as you go. I have found that labeling the functions f, f prime, g, and g prime can be an effective strategy for my students. It is helpful to read the problem first and then use the formula as a guideline for evaluating the derivative. This approach will be especially useful as the problems become more challenging. So if we call 2x squared, then 4x is the derivative of 2x squared. And f of x is g , and the derivative of f of x is fprime of x. Then, if you have written the words in that order, f is paired with g prime. The two outermost pieces are f and g. The innermost pieces are paired as follows: f with g, and g with h. Now you can simply write down what you see. The derivative of h prime of x-- again, make sure I use labels-- is equal to 2x squared times f prime of x plus 4x times f of x. Thus, I found h prime of 3 to be 2 times 3 squared times f prime of 3, or −5. Let's color-code this. I'll use purple.Times negative 5 plus 4 times 3 times f of 3.f of 3 is 4.So now that I have that, I can scroll down, do the math and find h prime of 3 is negative 42.