3 Derivatives 3.1 Differentiation Rules 3.3 Derivatives of Trigonometric Functions

3.2 Product and Quotient Rules

3.2.1 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 $\displaystyle f(x)=x^{2}$ , whose derivative is $\displaystyle f^{\prime}(x)=2x$ and not $\displaystyle\frac{d}{dx}(x)\cdot\frac{d}{dx}(x)=1\cdot 1=1$ .

Theorem 3.5 (Product Rule).

Let $\displaystyle f(x)$ and $\displaystyle g(x)$ be differentiable functions. Then

\displaystyle\frac{d}{dx}\left(f(x)g(x)\right)=\frac{d}{dx}(f(x))\cdot g(x)+% \frac{d}{dx}(g(x))\cdot f(x).

That is,

\displaystyle\text{ if }j(x)=f(x)g(x),\text{then}j^{\prime}(x)=f^{\prime}(x)g(% x)+g^{\prime}(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 $\displaystyle f(x)$ and $\displaystyle g(x)$ are differentiable functions. At a key point in this proof we need to use the fact that, since $\displaystyle g(x)$ is differentiable, it is also continuous. In particular, we use the fact that since $\displaystyle g(x)$ is continuous, $\displaystyle\lim_{h\to 0}g(x+h)=g(x)$ .

By applying the limit definition of the derivative to $\displaystyle j(x)=f(x)g(x)$ , we obtain

\displaystyle j^{\prime}(x)=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}.

By adding and subtracting $\displaystyle f(x)g(x+h)$ in the numerator, we have

\displaystyle j^{\prime}(x)=\lim_{h\to 0}\frac{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

\displaystyle j^{\prime}(x)=\lim_{h\to 0}\left(\frac{f(x+h)g(x+h)-f(x)g(x+h)}{% h}\right)+\lim_{h\to 0}\left(\frac{f(x)g(x+h)-f(x)g(x)}{h}\right).

Rearranging, we obtain

\displaystyle j^{\prime}(x)=\lim_{h\to 0}\left(\frac{f(x+h)-f(x)}{h}\cdot g(x+% h)\right)+\lim_{h\to 0}\left(\frac{g(x+h)-g(x)}{h}\cdot f(x)\right).

By using the continuity of $\displaystyle g(x)$ , the definition of the derivatives of $\displaystyle f(x)$ and $\displaystyle g(x)$ , and applying the limit laws, we arrive at the product rule,

\displaystyle j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x).

∎

Example 3.2.1 (Applying the Product Rule to Functions at a Point).

For $\displaystyle j(x)=f(x)g(x)$ , use the product rule to find $\displaystyle j^{\prime}(2)$ if $\displaystyle f(2)=3,f^{\prime}(2)=-4,g(2)=1$ , and $\displaystyle g^{\prime}(2)=6$ .

Solution: Since $\displaystyle j(x)=f(x)g(x),j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)$ , and hence

\displaystyle j^{\prime}(2)=f^{\prime}(2)g(2)+g^{\prime}(2)f(2)=(-4)(1)+(6)(3)% =14.

Example 3.2.2 (Applying the Product Rule to Binomials).

For $\displaystyle j(x)=(x^{2}+2)(3x^{3}-5x)$ , find $\displaystyle j^{\prime}(x)$ by applying the product rule. Check the result by first finding the product and then differentiating.

Solution: If we set $\displaystyle f(x)=x^{2}+2$ and $\displaystyle g(x)=3x^{3}-5x$ , then $\displaystyle f^{\prime}(x)=2x$ and $\displaystyle g^{\prime}(x)=9x^{2}-5$ . Thus,

\displaystyle j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)=(2x)(3x^{3}-5x% )+(9x^{2}-5)(x^{2}+2).

Simplifying, we have

\displaystyle j^{\prime}(x)=15x^{4}+3x^{2}-10.

To check, we see that $\displaystyle j(x)=3x^{5}+x^{3}-10x$ and, consequently, $\displaystyle j^{\prime}(x)=15x^{4}+3x^{2}-10$ .

Checkpoint 3.2.1.

Use the product rule to obtain the derivative of $\displaystyle j(x)=2x^{5}(4x^{2}+x)$ .

Hint: Set $\displaystyle f(x)=2x^{5}$ and $\displaystyle g(x)=4x^{2}+x$ and use the preceding example as a guide.

3.2.2 The Quotient Rule

Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that

\displaystyle\frac{d}{dx}(x^{2})=2x,\quad\text{not}\quad\frac{\frac{d}{dx}(x^{% 3})}{\frac{d}{dx}(x)}=\frac{3x^{2}}{1}=3x^{2}.

Theorem 3.6 (The Quotient Rule).

Let $\displaystyle f(x)$ and $\displaystyle g(x)$ be differentiable functions. Then

\displaystyle\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{\frac{d}{dx}(f(x% ))\cdot g(x)-\frac{d}{dx}(g(x))\cdot f(x)}{(g(x))^{2}}.

That is,

\displaystyle\text{ if }j(x)=\frac{f(x)}{g(x)},\text{then}j^{\prime}(x)=\frac{% f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^{2}}.

The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.

Example 3.2.3 (Applying the Quotient Rule).

Use the quotient rule to find the derivative of $\displaystyle k(x)=\frac{5x^{2}}{4x+3}$ .

Solution: Let $\displaystyle f(x)=5x^{2}$ and $\displaystyle g(x)=4x+3$ . Thus, $\displaystyle f^{\prime}(x)=10x$ and $\displaystyle g^{\prime}(x)=4$ . Substituting into the quotient rule, we have

\displaystyle k^{\prime}(x)=\frac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^% {2}}=\frac{10x(4x+3)-4(5x^{2})}{(4x+3)^{2}}.

Simplifying, we obtain

\displaystyle k^{\prime}(x)=\frac{20x^{2}+30x}{(4x+3)^{2}}.

Checkpoint 3.2.2.

Find the derivative of $\displaystyle h(x)=\frac{3x+1}{4x-3}$ .

Hint: Apply the quotient rule with $\displaystyle f(x)=3x+1$ and $\displaystyle g(x)=4x-3$ .

It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form $\displaystyle x^{k}$ where $\displaystyle k$ is a negative integer.

Theorem 3.7 (Extended Power Rule).

If $\displaystyle k$ is a negative integer, then

\displaystyle\frac{d}{dx}(x^{k})=kx^{k-1}.

Proof.

If $\displaystyle k$ is a negative integer, we may set $\displaystyle n=-k$ , so that $\displaystyle n$ is a positive integer with $\displaystyle k=-n$ . Since for each positive integer $\displaystyle n,x^{-n}=\frac{1}{x^{n}}$ , we may now apply the quotient rule by setting $\displaystyle f(x)=1$ and $\displaystyle g(x)=x^{n}$ . In this case, $\displaystyle f^{\prime}(x)=0$ and $\displaystyle g^{\prime}(x)=nx^{n-1}$ . Thus,

\displaystyle\frac{d}{\,dx}(x^{-n})=\frac{0(x^{n})-1(nx^{n-1})}{({x^{n}})^{2}}.

Simplifying, we see that

\displaystyle\frac{d}{\,dx}(x^{-n})=\frac{-nx^{n-1}}{x^{2n}}=-nx^{(n-1)-2n}=-% nx^{-n-1}.

Finally, observe that since $\displaystyle k=-n$ , by substituting we have

\displaystyle\frac{d}{dx}(x^{k})=kx^{k-1}.

∎

Example 3.2.4 (Using the Extended Power Rule).

Find $\displaystyle\frac{d}{dx}(x^{-4})$ .

Solution: By applying the extended power rule with $\displaystyle k=-4$ , we obtain

\displaystyle\frac{d}{dx}(x^{-4})=-4x^{-4-1}=-4x^{-5}.

Example 3.2.5 (Using the Extended Power Rule and the Constant Multiple Rule).

Use the extended power rule and the constant multiple rule to find the derivative of $\displaystyle f(x)=\frac{6}{x^{2}}$ .

Solution: It may seem tempting to use the quotient rule to find this derivative, and it would certainly not be incorrect to do so. However, it is far easier to differentiate this function by first rewriting it as $\displaystyle f(x)=6x^{-2}$ .

$\displaystyle f^{\prime}(x)$	$\displaystyle=\frac{d}{dx}\left(\frac{6}{x^{2}}\right)=\frac{d}{dx}\left(6x^{-% 2}\right)$	$\displaystyle\text{Rewrite }\frac{6}{x^{2}}\text{ as }6x^{-2}.$
	$\displaystyle=6\frac{d}{dx}(x^{-2})$	$\displaystyle\text{Apply the constant multiple rule}.$
	$\displaystyle=6(-2x^{-3})$	$\displaystyle\text{Use the extended power rule to differentiate}x^{-2}.$
	$\displaystyle=-12x^{-3}$	$\displaystyle\text{Simplify}.$

Checkpoint 3.2.3.

Find the derivative of $\displaystyle g(x)=\frac{1}{x^{7}}$ using the extended power rule.

Hint: Rewrite $\displaystyle g(x)=\frac{1}{x^{7}}=x^{-7}$ . Use the extended power rule with $\displaystyle k=-7$ .

3.2.3 Combining Differentiation Rules

As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one differentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we may find the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations of differentiation rules. A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in which we would evaluate the function.

Example 3.2.6 (Combining Differentiation Rules).

For $\displaystyle k(x)=3h(x)+x^{2}g(x)$ , find $\displaystyle k^{\prime}(x)$ .

Solution: Finding this derivative requires the sum rule, the constant multiple rule, and the product rule.

$\displaystyle k^{\prime}(x)$	$\displaystyle=\frac{d}{dx}\left(3h(x)+x^{2}g(x)\right)=\frac{d}{dx}\left(3h(x)% \right)+\frac{d}{dx}\left(x^{2}g(x)\right)$	$\displaystyle\text{Apply the sum rule}.$
	$\displaystyle=3\frac{d}{dx}(h(x))+\left(\frac{d}{dx}(x^{2})g(x)+\frac{d}{dx}(g% (x))x^{2}\right)$	$\displaystyle\begin{aligned} \text{Apply the constant multiple rule to}\\ \text{differentiate}3h(x)\text{and the product}\\ \text{rule to differentiate}x^{2}g(x).\end{aligned}$
	$\displaystyle=3h^{\prime}(x)+2xg(x)+g^{\prime}(x)x^{2}$

Example 3.2.7 (Extending the Product Rule).

For $\displaystyle k(x)=f(x)g(x)h(x)$ , express $\displaystyle k^{\prime}(x)$ in terms of $\displaystyle f(x),g(x),h(x)$ , and their derivatives.

Solution: We can think of the function $\displaystyle k(x)$ as the product of the function $\displaystyle f(x)g(x)$ and the function $\displaystyle h(x)$ . That is, $\displaystyle k(x)=(f(x)g(x))\cdot h(x)$ . Thus,

$\displaystyle k^{\prime}(x)$	$\displaystyle=\frac{d}{dx}\left(f(x)g(x)\right)\cdot h(x)+\frac{d}{dx}(h(x))% \cdot\left(f(x)g(x)\right)$	Apply the product rule to the product of $\displaystyle f(x)g(x)$ and $\displaystyle h(x)$ .
	$\displaystyle=(f^{\prime}(x)g(x)+g^{\prime}(x)f(x))h(x)+h^{\prime}(x)f(x)g(x)$	$\displaystyle\text{Apply the product rule to }f(x)g(x).$
	$\displaystyle=f^{\prime}(x)g(x)h(x)+f(x)g^{\prime}(x)h(x)+f(x)g(x)h^{\prime}(x).$	$\displaystyle\text{Simplify}.$

Example 3.2.8 (Combining the Quotient Rule and the Product Rule).

For $\displaystyle h(x)=\frac{2x^{3}k(x)}{3x+2}$ , find $\displaystyle h^{\prime}(x)$ .

Solution: This procedure is typical for finding the derivative of a rational function.

$\displaystyle\displaystyle h^{\prime}(x)$	$\displaystyle\displaystyle=\frac{\frac{d}{dx}\left(2x^{3}k(x)\right)\cdot(3x+2% )-\frac{d}{dx}(3x+2)\cdot\left(2x^{3}k(x)\right)}{(3x+2)^{2}}$	$\displaystyle\displaystyle\text{Apply the quotient rule}.$
	$\displaystyle\displaystyle=\frac{(6x^{2}k(x)+k^{\prime}(x)\cdot 2x^{3})(3x+2)-% 3(2x^{3}k(x))}{(3x+2)^{2}}$
	$\displaystyle\displaystyle=\frac{-6x^{3}k(x)+18x^{3}k(x)+12x^{2}k(x)+6x^{4}k^{% \prime}(x)+4x^{3}k^{\prime}(x)}{(3x+2)^{2}}$	$\displaystyle\displaystyle\text{Simplify}.$

Checkpoint 3.2.4.

Find $\displaystyle\frac{d}{dx}\left(3f(x)-2g(x)\right)$ .

Hint: Apply the difference rule and the constant multiple rule.

Example 3.2.9 (Determining Where a Function Has a Horizontal Tangent).

Determine the values of $\displaystyle x$ for which $\displaystyle f(x)=x^{3}-7x^{2}+8x+1$ has a horizontal tangent line.

Solution: To find the values of $\displaystyle x$ for which $\displaystyle f(x)$ has a horizontal tangent line, we must solve $\displaystyle f^{\prime}(x)=0$ . Since

\displaystyle f^{\prime}(x)=3x^{2}-14x+8=(3x-2)(x-4),

we must solve $\displaystyle(3x-2)(x-4)=0$ . Thus we see that the function has horizontal tangent lines at $\displaystyle x=\frac{2}{3}$ and $\displaystyle x=4$ as shown in the following graph.

Figure 3.5: This function has horizontal tangent lines at

\displaystyle x

= 2/3 and

\displaystyle x

= 4.

Example 3.2.10 (Finding a Velocity).

The position of an object on a coordinate axis at time $\displaystyle t$ is given by $\displaystyle s(t)=\frac{t}{t^{2}+1}$ . What is the initial velocity of the object?

Solution: Since the initial velocity is $\displaystyle v(0)=s^{\prime}(0)$ , begin by finding $\displaystyle s^{\prime}(t)$ by applying the quotient rule:

\displaystyle s^{\prime}(t)=\frac{1(t^{2}+1)-2t(t)}{(t^{2}+1)^{2}}=\frac{1-t^{% 2}}{(t^{2}+1)^{2}}.

After evaluating, we see that $\displaystyle v(0)=1$ .

Checkpoint 3.2.5.

Find the values of $\displaystyle x$ for which the graph of $\displaystyle f(x)=4x^{2}-3x+2$ has a tangent line parallel to the line $\displaystyle y=2x+3$ .

Hint: Solve $\displaystyle f^{\prime}(x)=2$ .

Key Concepts

•

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.
•

The derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function.
•

We used the limit definition of the derivative to develop formulas that allow us to find derivatives without resorting to the definition of the derivative. These formulas can be used singly or in combination with each other.