The Product Rule
Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this pattern. To see why we cannot use this pattern, consider the function (f(x)=x^2), whose derivative is (f′(x)=2x) and not (dfrac{d}{dx}(x)⋅dfrac{d}{dx}(x)=1⋅1=1.)
So for example if I have some function F of X and it can be expressed as the quotient of two expressions. So let's say U of X over V of X. Then the quotient rule tells us that F prime of X is going to be equal to and this is going to look a little bit complicated but once we apply it, you'll hopefully get a little bit more comfortable with it. Quotient Rule Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture. The quotient rule is used to determine the derivative of one function divided by another. 3.2 The Product Rule and the Quotient Rule Math 1271, TA: Amy DeCelles 1. Overview You need to memorize the product rule and the quotient rule. And, more than that actually: you need to internalize them. The best way to do that is just by practicing until you can use them without even thinking about it. Product Rule: (fg)0= f0g + fg0. AP Calculus AB Notes, Worksheets and Classroom Policies. Section 2.3a The Product and Quotient Rules and Higher-Order Derivatives.pdf View Download.
Product Rule
Let (f(x)) and (g(x)) be differentiable functions. Then
[dfrac{d}{dx}(f(x)g(x))=dfrac{d}{dx}(f(x))⋅g(x)+dfrac{d}{dx}(g(x))⋅f(x).]
That is,
[if j(x)=f(x)g(x),thenj′(x)=f′(x)g(x)+g′(x)f(x).]
This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.
Proof
We begin by assuming that (f(x)) and (g(x)) are differentiable functions. At a key point in this proof we need to use the fact that, since (g(x)) is differentiable, it is also continuous. In particular, we use the fact that since (g(x)) is continuous, (lim_{h→0}g(x+h)=g(x).)
By applying the limit definition of the derivative to ((x)=f(x)g(x),) we obtain
[j′(x)=lim_{h→0}dfrac{f(x+h)g(x+h)−f(x)g(x)}{h}.]
By adding and subtracting (f(x)g(x+h)) in the numerator, we have
3.2 Quotient Ruleap Calculus Calculator
[j′(x)=lim_{h→0}dfrac{f(x+h)g(x+h)−f(x)g(x+h)+f(x)g(x+h)−f(x)g(x)}{h}.]
After breaking apart this quotient and applying the sum law for limits, the derivative becomes
[j′(x)=lim_{h→0}dfrac{(f(x+h)g(x+h)−f(x)g(x+h)}{h})+lim_{h→0}dfrac{(f(x)g(x+h)−f(x)g(x)}{h}.]
Rearranging, we obtain
[j′(x)=lim_{h→0}dfrac{(f(x+h)−f(x)}{h}⋅g(x+h))+lim_{h→0}(dfrac{g(x+h)−g(x)}{h}⋅f(x)).]
By using the continuity of (g(x)), the definition of the derivatives of (f(x)) and (g(x)), and applying the limit laws, we arrive at the product rule,
[j′(x)=f′(x)g(x)+g′(x)f(x).]
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Example (PageIndex{7}): Applying the Product Rule to Constant Functions
For (j(x)=f(x)g(x)), use the product rule to find (j′(2)) if (f(2)=3,f′(2)=−4,g(2)=1), and (g′(2)=6).
Solution
Since (j(x)=f(x)g(x),j′(x)=f′(x)g(x)+g′(x)f(x),) and hence
[j′(2)=f′(2)g(2)+g′(2)f(2)=(−4)(1)+(6)(3)=14.]
Example (PageIndex{8}): Applying the Product Rule to Binomials
For (j(x)=(x^2+2)(3x^3−5x),) find (j′(x)) by applying the product rule. Check the result by first finding the product and then differentiating.
Solution
If we set (f(x)=x^2+2) and (g(x)=3x^3−5x), then (f′(x)=2x) and (g′(x)=9x^2−5). Thus,
(j′(x)=f′(x)g(x)+g′(x)f(x)=(2x)(3x^3−5x)+(9x^2−5)(x^2+2).)
Simplifying, we have
[j′(x)=15x^4+3x^2−10.]
To check, we see that (j(x)=3x^5+x^3−10x) and, consequently, (j′(x)=15x^4+3x^2−10.)
Exercise (PageIndex{6})
Use the product rule to obtain the derivative of [j(x)=2x^5(4x^2+x).]
3.2 Quotient Ruleap Calculus Solver
Set (f(x)=2x^5) and (g(x)=4x^2+x) and use the preceding example as a guide.
[j′(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5.]