2.4 Product and Quotient Rules
2.4 Product and Quotient Rules
The basic rules let us tackle simple functions produced from basic types of functions through addition/subtraction and multiplication by a constant.
In this section we will discuss the rules that allow us to find the derivatives of functions produced from multiplication and division of functions, not just by a constant.
Example 1
Find the derivative of
[latex]h(x)=\left(4x^3-11\right)(x+3)[/latex]
Answer: [latex]h'(x)=?[/latex]
This function is not a simple sum or difference of polynomials. It’s a product of polynomials. We can simply multiply it out to find its derivative:
[latex]\begin{align*} h(x) & = \left(4x^3-11\right)(x+3)\\ & = 4x^4-11x+12x^3-33\\ &= 4x^4+12x^3-11x-33\\ \end{align*}[/latex]
Analyzing the function we see that
[latex]h(x)=\overbrace{\underbrace{4\cdot \underbrace{x^4}_{power}}_{const.\ mult.}+\underbrace{12\cdot \underbrace{x^3}_{power}}_{const.\ mult.}-\underbrace{11\cdot \underbrace{x}_{lin.}}_{const.\ mult.}-\underbrace{33}_{const.}}^{sum/difference}[/latex]
Thus
[latex]h'(x)= 4\cdot 4x^{4-1}+12\cdot 3x^{3-1}-11\cdot 1-0=16x^3+36x^2-11[/latex]
Now suppose we wanted to find the derivative of
[latex]f(x)=\left(4x^5+x^3-1.5x^2-11\right)\left(x^7-7.25x^5+120x+3\right)[/latex]
This function is not a simple sum or difference of polynomials. It’s a product of polynomials. We could ‘simply’ multiply it out to find its derivative as before – who wants to volunteer? Nobody?
We’ll need a rule for finding the derivative of a product so we don’t have to multiply everything out. It would be great if we can just take the derivatives of the factors and multiply them, but unfortunately that won’t give the right answer.
A classic error students make with products of functions is to calculate the derivative of each individual function and then multiply the derivatives. It turns out that this is incorrect!
The rules for finding derivatives of products and quotients are a little complicated, but they save us the much more complicated algebra we might face if we were to try to multiply things out. They also let us deal with products where the factors are not polynomials. We can use these rules, together with the basic rules, to find derivatives of many complicated looking functions.
Video Demonstration
Products of functions; Quotients of functions
© 2014 Eric Bancroft
Derivative Rules: Product and Quotient Rules
In what follows, we assume that [latex]f[/latex] and [latex]g[/latex] are differentiable functions with respect to [latex]x[/latex].
Product Rule
The derivative of a product of two functions is the derivative of the first times the original second, plus the original first times the derivative of the second.
[latex]\dfrac{d}{dx}\left( f\cdot g \right)=f'\cdot g+f\cdot g'[/latex]
The product rule can extend to a product of several functions; the pattern continues – take the derivative of each factor in turn, multiplied by all the other factors left alone, and add them up:
[latex]\dfrac{d}{dx}\left( f\cdot g\cdot h \right)=f'\cdot g\cdot h+f\cdot g'\cdot h+f\cdot g\cdot h'[/latex]
Quotient Rule
The derivative of a quotient of two functions is the derivative of the numerator times the original denominator, minus the original numerator times the derivative of the denominator, all divided by the square of the denominator.
[latex]\dfrac{d}{dx}\left( \frac{f}{g} \right)=\dfrac{f'\cdot g-f\cdot g'}{g^2}[/latex]
The numerator of the result resembles the product rule, but there is a minus instead of a plus; the minus sign goes with the [latex]g’[/latex]. The denominator is simply the square of the original denominator – no derivatives there.
Example 2
Find the derivative of
[latex]h(x)=\left(4x^3-11\right)(x+3)[/latex]
Answer: [latex]h'(x)=?[/latex]
This is the same function we found the derivative of in Example 1, but let’s use the product rule and check to see if we get the same answer.
Our detailed analysis shows us that:
[latex]h(x)=\overbrace{\overbrace{(\underbrace{4\cdot \underbrace{x^3}_{power\ fn.}}_{const.\ mult.}-\underbrace{11}_{const.}}^{difference})(\overbrace{\underbrace{x}_{lin.}+\underbrace{3}_{const}}^{sum})}^{product}[/latex]
Using the product rule, the constant multiple rule, the power rule, the constant rule, and the sum and difference rules, if we name the first and the second function in the product by [latex]F[/latex] and [latex]S[/latex], we get
[latex]\begin{align*} h'(x) & = F'\cdot S+F\cdot S'=(4\cdot 3x^{3-1}-0)\cdot (x+3)+(4x^3-11)\cdot (1+0) \\\\ & = \left(12x^2\right)(x+3)+\left(4x^3-11\right)(1)=12x^3+36x^2+4x^3-11 \\\\ & = 16x^3+36x^2-11 \end{align*}[/latex]
Example 3
Find the derivative of [latex]m(t)=e^t\ln(t)[/latex].
Answer: [latex]\frac{dm}{dt}=?[/latex]
This is a product of a natural exponential function and the natural logarithmic function, so we need to use the product rule and the rules for the exponential and logarithmic functions. For easier recall of the rule, let’s call the first function [latex]F[/latex] and the second function [latex]S[/latex].
Then
[latex]\dfrac{dm}{dt}=F'\cdot S+F\cdot S'=\left(e^t\right)\left(\ln(t)\right)+\left(e^t\right)\left(\frac{1}{t}\right)=e^t\ln(t)+\frac{e^t}{t}[/latex]
Note that the function in the last example was one we could not have done by “multiplying out” because functions being multiplied are of different type.
Example 4
Find the derivative of [latex]y=\frac{x^4+4^x}{3+16x^3}[/latex].
Answer:
[latex]y'=?[/latex]
[latex]y=\overbrace{\dfrac{\overbrace{\overbrace{x^4}^{power fn.}+\overbrace{4^x}^{exp. fn.}}^{sum}}{\underbrace{\underbrace{3}_{const.}+\underbrace{16\cdot \underbrace{x^3}_{power fn.}}_{const. mult.}}_{sum}}}^{quotient}[/latex]
Therefore
[latex]\begin{align*} y'&=\dfrac{T'\cdot B-T\cdot B'}{B^2}\\\\ &=\dfrac{\left(4\cdot x^{4-1}+\ln(4)\cdot 4^x \right)\left(3+16x^3 \right)-\left(x^4+4^x \right)\left(0+16\cdot 3x^{3-1} \right)}{\left(3+16x^3 \right)^2}\\\\ &=\dfrac{\left(4x^3 +\ln(4)\cdot 4^x\right)\left(3+16x^3 \right)-\left(x^4+4^x \right)\left(48x^2 \right)}{\left(3+16x^3 \right)^2} \end{align*}[/latex]
Don’t worry about simplifying that! Remember that simple depends on what you will do next; in this case, we were asked to find the derivative, and we’ve done that. You should do any basic simplifications, such as multiplying constants together or doing obvious cancellations or combining of terms, but otherwise don’t spend too much time simplifying unless there is a reason to simplify further.
Video Demonstration
Nine more examples
© 2014 Eric Bancroft
Example 5
The sigmoid function [latex]\sigma(x)[/latex] is an S-shaped function that plays a large role in neural networks in classification problems and is used in so-called logistic regression to map any real-valued number into a range of 0 to 1. It is defined as
[latex]\sigma(x)=\dfrac{e^x}{1+e^x}[/latex]
Find [latex]\sigma'(x)[/latex].
Answer: [latex]\sigma'(x)=?[/latex]
Let’s do the analysis of the function:
[latex]\sigma(x)=\overbrace{\dfrac{\overbrace{e^x}^{nat.\ exp.\ fn.}}{\underbrace{\underbrace{1}_{const.}+\underbrace{e^x}_{nat. exp. fn}}_{sum}}}^{quotient}[/latex]
Therefore,
[latex]\begin{align*} \sigma'(t) & = \dfrac{e^x(1+e^x)-e^x(0+e^x)}{\left(1+e^x\right)^2} \\ & = \dfrac{e^x}{\left(1+e^x\right)^2} \end{align*}[/latex]
Section Exercises
Work on the following exercises. Discuss your solutions with your peers and/or course instructor.
Section 2.4 Exercises – Product and Quotient Rules
