1.1: Product and Quotient Rule

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 \(\dfrac{d}{dx}(x)⋅\dfrac{d}{dx}(x)=1⋅1=1.\)

Product Rule

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

\[\dfrac{d}{dx}(f(x)g(x))=\dfrac{d}{dx}(f(x))⋅g(x)+\dfrac{d}{dx}(g(x))⋅f(x). \nonumber \]

That is,

\[\text{if }p(x)=f(x)g(x),\quad \text{then }p′(x)=f′(x)g(x)+g′(x)f(x).\nonumber \]

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, \(\displaystyle \lim_{h→0}g(x+h)=g(x).\)

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

\[ p′(x)=\lim_{h→0}\dfrac{f(x+h)g(x+h)−f(x)g(x)}{h}.\nonumber \]

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

\[p′(x)=\lim_{h→0}\dfrac{f(x+h)g(x+h)−f(x)g(x+h)+f(x)g(x+h)−f(x)g(x)}{h}.\nonumber \]

After breaking apart this quotient and applying the sum law for limits, the derivative becomes

\[p′(x)=\lim_{h→0}\dfrac{f(x+h)g(x+h)−f(x)g(x+h)}{h}+\lim_{h→0}\dfrac{f(x)g(x+h)−f(x)g(x)}{h}.\nonumber \]

Rearranging, we obtain

\[\begin{align*}p′(x)&=\lim_{h→0}\left(\dfrac{f(x+h)−f(x)}{h}⋅g(x+h)\right)+\lim_{h→0}\left(\dfrac{g(x+h)−g(x)}{h}⋅f(x)\right)\\[4pt]
&= \left(\lim_{h→0}\dfrac{f(x+h)−f(x)}{h}\right)⋅\left(\lim_{h→0}\;g(x+h)\right)+\left(\lim_{h→0}\dfrac{g(x+h)−g(x)}{h}\right)⋅f(x)\end{align*}\]

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,

\[p′(x)=f′(x)g(x)+g′(x)f(x).\nonumber \]

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Example \(\PageIndex{7}\): Applying the Product Rule to Constant Functions

For \(p(x)=f(x)g(x)\), use the product rule to find \(p′(2)\) if \(f(2)=3,\; f′(2)=−4,\; g(2)=1\), and \(g′(2)=6\).

Solution

Since \(p(x)=f(x)g(x)\), \(p′(x)=f′(x)g(x)+g′(x)f(x),\) and hence

\(p′(2)=f′(2)g(2)+g′(2)f(2)=(−4)(1)+(6)(3)=14.\)

Example \(\PageIndex{8}\): Applying the Product Rule to Binomials

For \(p(x)=(x^2+2)(3x^3−5x),\) find \(p′(x)\) 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,

\(p′(x)=f′(x)g(x)+g′(x)f(x)=(2x)(3x^3−5x)+(9x^2−5)(x^2+2).\)

Simplifying, we have

\[p′(x)=15x^4+3x^2−10. \nonumber \]

To check, we see that \(p(x)=3x^5+x^3−10x\) and, consequently, \(p′(x)=15x^4+3x^2−10.\)

Exercise \(\PageIndex{6}\)

Use the product rule to obtain the derivative of \(p(x)=2x^5(4x^2+x).\)

Hint

Set \(f(x)=2x^5\) and \(g(x)=4x^2+x\) and use the preceding example as a guide.

Answer

\(p′(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5.\)

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

\[\dfrac{d}{dx}(x^2)=2x,\text{ not }\dfrac{\dfrac{d}{dx}(x^3)}{\dfrac{d}{dx}(x)}=\dfrac{3x^2}{1}=3x^2.\nonumber \]

The Quotient Rule

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

\[\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{\dfrac{d}{dx}(f(x))⋅g(x)−\dfrac{d}{dx}(g(x))⋅f(x)}{\big(g(x)\big)^2}. \nonumber \]

That is, if

\[q(x)=\dfrac{f(x)}{g(x)}\nonumber \]

then

\[q′(x)=\dfrac{f′(x)g(x)−g′(x)f(x)}{\big(g(x)\big)^2}.\nonumber \]

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 \(\PageIndex{9}\): Applying the Quotient Rule

Use the quotient rule to find the derivative of \(q(x)=\dfrac{5x^2}{4x+3}.\)

Solution

Let \(f(x)=5x^2\) and \(g(x)=4x+3\). Thus, \(f′(x)=10x\) and \(g′(x)=4\).

Substituting into the quotient rule, we have

\[q′(x)=\dfrac{f′(x)g(x)−g′(x)f(x)}{(g(x))^2}=\dfrac{10x(4x+3)−4(5x^2)}{(4x+3)^2}.\nonumber \]

Simplifying, we obtain

\[q′(x)=\dfrac{20x^2+30x}{(4x+3)^2}\nonumber \]

Exercise \(\PageIndex{7}\)

Find the derivative of \(h(x)=\dfrac{3x+1}{4x−3}\).

Hint

Apply the quotient rule with \(f(x)=3x+1\) and \(g(x)=4x−3\).

Answer

\(h′(x)=−\dfrac{13}{(4x−3)^2}.\)

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

Extended Power Rule

If \(k\) is a negative integer, then

\[\dfrac{d}{dx}(x^k)=kx^{k−1}. \nonumber \]

Proof

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

\[\dfrac{d}{dx}(x^{−n})=\dfrac{0(x^n)−1(nx^{n−1})}{(x^n)^2}.\nonumber \]

Simplifying, we see that

\[\begin{align*} \dfrac{d}{dx}(x^{−n}) &=\dfrac{−nx^{n−1}}{x^{2n}}\\[4pt]&=−nx^{(n−1)−2n}\\[4pt]&=−nx^{−n−1}.\end{align*}\]

Finally, observe that since \(k=−n\), by substituting we have

\[\dfrac{d}{dx}(x^k)=kx^{k−1}.\nonumber \]

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Example \(\PageIndex{10}\): Using the Extended Power Rule

Find \(\dfrac{d}{dx}(x^{−4})\).

Solution

By applying the extended power rule with \(k=−4\), we obtain

\[\dfrac{d}{dx}(x^{−4})=−4x^{−4−1}=−4x^{−5}.\nonumber \]

Example \(\PageIndex{11}\): Using the Extended Power Rule and the Constant Multiple Rule

Use the extended power rule and the constant multiple rule to find \(f(x)=\dfrac{6}{x^2}\).

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 \(f(x)=6x^{−2}\).

\(\begin{align*} f′(x)&=\dfrac{d}{dx}\left(\dfrac{6}{x^2}\right)=\dfrac{d}{dx}\left(6x^{−2}\right) & & \text{Rewrite }\dfrac{6}{x^2}\text{ as }6x^{−2}.\\[4pt]
&=6\dfrac{d}{dx}\left(x^{−2}\right) & & \text{Apply the constant multiple rule.}\\[4pt]
&=6(−2x^{−3}) & & \text{Use the extended power rule to differentiate }x^{−2}.\\[4pt]
&=−12x^{−3} & & \text{Simplify.} \end{align*} \)

Exercise \(\PageIndex{8}\)

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

Hint

Rewrite \(g(x)=\dfrac{1}{x^7}=x^{−7}\). Use the extended power rule with \(k=−7\).

Answer

\(g′(x)=−7x^{−8}\).