2.4 The Product and Quotient Rules

The previous section showed that, in some ways, derivatives behave nicely. The Constant Multiple and Sum/Difference Rules established that the derivative of $f(x)=5x^2+\sin x$ was not complicated. We neglected computing the derivative of things like $g(x)=5x^2\sin x$ and $h(x)=\frac{5x^2}{\sin x}$ on purpose; their derivatives are not as straightforward. (If you had to guess what their respective derivatives are, you would probably guess wrong.) For these, we need the Product and Quotient Rules, respectively, which are defined in this section.

We begin with the Product Rule.

Important: ${\displaystyle \frac{d}{dx}}\left(f(x)g(x)\right)\ne f^{\prime }(x)g^{\prime }(x)$. While this answer is simpler than the Product Rule, it is wrong. We can show that this is wrong by considering
$f(x)=x^2$ and $g(x)=x^5$.
Using the WRONG rule we get ${\displaystyle \frac{d}{dx}}[f(x)g(x)]=2x\cdot 5x^4=10x^5$. However, when we simplify the product first and apply the Power Rule, $f\cdot g=x^2\cdot x^5=x^7$ and

Applying the real Product Rule we see that,

We practice using this new rule in an example, followed by a proof of the theorem.

• Proof of the Product Rule

  
  By the limit definition, we have

  $$\frac{d}{dx}\left(f(x)g(x)\right)=\underset{h\to 0}{lim}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}.$$

  We now do something a bit unexpected; add 0 to the numerator (so that nothing is changed) in the form of $-f(x+h)g(x)+f(x+h)g(x)$, and then do some regrouping as shown.

  	${\displaystyle \frac{d}{dx}}$	$\left(f(x)g(x)\right)$
  		$=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)g(x+h)-f(x)g(x)}{h}}\mathit{\hspace{1em}\hspace{1em}}\text{(now add 0 to the numerator)}$
  		$=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}}$
  		$=\underset{h\to 0}{lim}{\displaystyle \frac{\left(f(x+h)g(x+h)-f(x+h)g(x)\right)+\left(f(x+h)g(x)-f(x)g(x)\right)}{h}}$
  		$=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)g(x+h)-f(x+h)g(x)}{h}}+\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)g(x)-f(x)g(x)}{h}}$
  		$=\underset{h\to 0}{lim}f(x+h){\displaystyle \frac{g(x+h)-g(x)}{h}}+\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}g(x)$
  		$=\underset{h\to 0}{lim}f(x+h)\underset{h\to 0}{lim}{\displaystyle \frac{g(x+h)-g(x)}{h}}+\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}\underset{h\to 0}{lim}g(x)$
  		$=f(x)g^{\prime }(x)+f^{\prime }(x)g(x).\mathit{∎}$

It is often true that we can recognize that a theorem is true through its proof yet somehow doubt its applicability to real problems. In the following example, we compute the derivative of a product of functions in two ways to verify that the Product Rule is indeed “right.”

We consider one more example before discussing another derivative rule.

We have learned how to compute the derivatives of sums, differences, and products of functions. We now learn how to find the derivative of a quotient of functions.

• Proof of the Quotient Rule

  
  Let the functions $f$ and $g$ be defined and $g(x)\ne 0$ on an open interval $I$. By the definition of derivative,

  	${\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{f(x)}{g(x)}}\right)$	$=\underset{h\to 0}{lim}{\displaystyle \frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}}$
  		$=\underset{h\to 0}{lim}\left[\left({\displaystyle \frac{f(x+h)}{g(x+h)}}-{\displaystyle \frac{f(x)}{g(x)}}\right)\cdot {\displaystyle \frac{1}{h}}\right]$
  		$=\underset{h\to 0}{lim}\left[\left({\displaystyle \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}}\right)\cdot {\displaystyle \frac{1}{h}}\right]$

  Adding and subtracting the term $f(x)g(x)$ in the numerator does not change the value of the expression and allows us to separate $f$ and $g$ so that
  

  	${\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}\right)$	$=\underset{h\to 0}{lim}\left[\left({\displaystyle \frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x\right)+f\left(x\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)}{g\left(x+h\right)g\left(x\right)}}\right)\cdot {\displaystyle \frac{1}{h}}\right]$
  		$=\underset{h\to 0}{lim}\left[{\displaystyle \frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x\right)}{hg\left(x+h\right)g\left(x\right)}}+{\displaystyle \frac{f\left(x\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)}{hg\left(x+h\right)g\left(x\right)}}\right]$
  		$=\underset{h\to 0}{lim}\left[g\left(x\right){\displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{hg\left(x+h\right)g\left(x\right)}}+f\left(x\right){\displaystyle \frac{g\left(x\right)-g\left(x+h\right)}{hg\left(x+h\right)g\left(x\right)}}\right]$
  		$=\underset{h\to 0}{lim}{\displaystyle \frac{g\left(x\right)\frac{f\left(x+h\right)-f\left(x\right)}{h}-f\left(x\right)\frac{g\left(x+h\right)-g\left(x\right)}{h}}{g\left(x+h\right)g\left(x\right)}}$
  		$={\displaystyle \frac{\underset{h\to 0}{lim}g\left(x\right)\cdot \underset{h\to 0}{lim}{\displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h}}-\underset{h\to 0}{lim}f\left(x\right)\cdot \underset{h\to 0}{lim}{\displaystyle \frac{g\left(x+h\right)-g\left(x\right)}{h}}}{\underset{h\to 0}{lim}g\left(x+h\right)\cdot \underset{h\to 0}{lim}g\left(x\right)}}$
  		$={\displaystyle \frac{g\left(x\right)f^{\prime }\left(x\right)-f\left(x\right)g^{\prime }\left(x\right)}{{\left[g\left(x\right)\right]}^2}}\mathit{∎}$

Let’s practice using the Quotient Rule.

The Quotient Rule allows us to fill in holes in our understanding of derivatives of the common trigonometric functions. We start with finding the derivative of the tangent function.

We include this result in the following theorem about the derivatives of the trigonometric functions. Recall we found the derivative of $y=\sin x$ in Example 2.1.6 and stated the derivative of the cosine function in Theorem 2.3.1. The derivatives of the cotangent, cosecant and secant functions can all be computed directly using Theorem 2.3.1 and the Quotient Rule.

The proofs of these derivatives have been presented or left as exercises. To remember the above, it may be helpful to keep in mind that the derivatives of the trigonometric functions that start with “c” have a minus sign in them.

When we stated the Power Rule in Section 2.3 we claimed that it worked for all $n\in ℝ$ but only provided the proof for non-negative integers. The next example uses the Quotient Rule to provide justification of the Power Rule for $n\in \mathbb{Z}$.

The derivative of $y={\displaystyle \frac{1}{x^n}}$ turned out to be rather nice. It gets better. Consider:

Thus, for all $n\in \mathbb{Z}$, we can officially apply the Power Rule: multiply by the power, then subtract 1 from the power.

Taking the derivative of many functions is relatively straightforward. It is clear (with practice) what rules apply and in what order they should be applied. Other functions present multiple paths; different rules may be applied depending on how the function is treated. One of the beautiful things about calculus is that there is not “the” right way; each path, when applied correctly, leads to the same result, the derivative. We demonstrate this concept in an example.

Example 2.4.9 demonstrates three methods of finding $f^{\prime }$. One is hard pressed to argue for a “best method” as all three gave the same result without too much difficulty, although it is clear that using the Product Rule required more steps. Ultimately, the important principle to take away from this is: simplify the answer to a form that seems “simple” and easy to interpret. They are equal; they are all correct. The most appropriate form of $f^{\prime }$ depends on what we need to do with the function next. For later problems it will be important for us to determine the most appropriate form to use and to move flexibly between the different forms.

In the next section we continue to learn rules that allow us to more easily compute derivatives than using the limit definition directly. We have to memorize the derivatives of a certain set of functions, such as “the derivative of $\sin x$ is $\cos x$.” The Sum/Difference, Constant Multiple, Power, Product and Quotient Rules show us how to find the derivatives of certain combinations of these functions. The next section shows how to find the derivatives when we compose these functions together.