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

\[\frac{d}{d x} \left(\right. x^{2} \left.\right) = 2 x , \text{not} \frac{\frac{d}{d x} \left(\right. x^{3} \left.\right)}{\frac{d}{d x} \left(\right. x \left.\right)} = \frac{3 x^{2}}{1} = 3 x^{2} .\]

Theorem 3.6

The Quotient Rule

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

\[\frac{d}{d x} \left(\right. \frac{f \left(\right. x \left.\right)}{g \left(\right. x \left.\right)} \left.\right) = \frac{\frac{d}{d x} \left(\right. f \left(\right. x \left.\right) \left.\right) \cdot g \left(\right. x \left.\right) - \frac{d}{d x} \left(\right. g \left(\right. x \left.\right) \left.\right) \cdot f \left(\right. x \left.\right)}{\left(\left(\right. g \left(\right. x \left.\right) \left.\right)\right)^{2}} .\]

That is,

\[\text{if} j \left(\right. x \left.\right) = \frac{f \left(\right. x \left.\right)}{g \left(\right. x \left.\right)} , \text{then} j^{'} \left(\right. x \left.\right) = \frac{f^{'} \left(\right. x \left.\right) g \left(\right. x \left.\right) - g^{'} \left(\right. x \left.\right) f \left(\right. x \left.\right)}{\left(\left(\right. g \left(\right. x \left.\right) \left.\right)\right)^{2}} .\]

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 3.25

Applying the Quotient Rule

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

Solution

Let \(f \left(\right. x \left.\right) = 5 x^{2}\) and \(g \left(\right. x \left.\right) = 4 x + 3 .\) Thus, \(f^{'} \left(\right. x \left.\right) = 10 x\) and \(g^{'} \left(\right. x \left.\right) = 4 .\) Substituting into the quotient rule, we have

\[k^{'} \left(\right. x \left.\right) = \frac{f^{'} \left(\right. x \left.\right) g \left(\right. x \left.\right) - g^{'} \left(\right. x \left.\right) f \left(\right. x \left.\right)}{\left(\left(\right. g \left(\right. x \left.\right) \left.\right)\right)^{2}} = \frac{10 x \left(\right. 4 x + 3 \left.\right) - 4 \left(\right. 5 x^{2} \left.\right)}{\left(\left(\right. 4 x + 3 \left.\right)\right)^{2}} .\]

Simplifying, we obtain

\[k^{'} \left(\right. x \left.\right) = \frac{20 x^{2} + 30 x}{\left(\left(\right. 4 x + 3 \left.\right)\right)^{2}} .\]

Checkpoint 3.17

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

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.

Theorem 3.7

Extended Power Rule

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

\[\frac{d}{d x} \left(\right. x^{k} \left.\right) = k x^{k - 1} .\]

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 \(n , x^{− n} = \frac{1}{x^{n}} ,\) we may now apply the quotient rule by setting \(f \left(\right. x \left.\right) = 1\) and \(g \left(\right. x \left.\right) = x^{n} .\) In this case, \(f^{'} \left(\right. x \left.\right) = 0\) and \(g^{'} \left(\right. x \left.\right) = n x^{n - 1} .\) Thus,

\[\frac{d}{d x} \left(\right. x^{− n} \left.\right) = \frac{0 \left(\right. x^{n} \left.\right) - 1 \left(\right. n x^{n - 1} \left.\right)}{\left(\right. x^{n} \left.\right)^{2}} .\]

Simplifying, we see that

\[\frac{d}{d x} \left(\right. x^{− n} \left.\right) = \frac{− n x^{n - 1}}{x^{2 n}} = − n x^{\left(\right. n - 1 \left.\right) - 2 n} = − n x^{− n - 1} .\]

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

\[\frac{d}{d x} \left(\right. x^{k} \left.\right) = k x^{k - 1} .\]

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Example 3.26

Using the Extended Power Rule

Find \(\frac{d}{d x} \left(\right. x^{−4} \left.\right) .\)

Solution

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

\[\frac{d}{d x} \left(\right. x^{−4} \left.\right) = −4 x^{−4 - 1} = −4 x^{−5} .\]

Example 3.27

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 \(f \left(\right. x \left.\right) = \frac{6}{x^{2}} .\)

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 \(f \left(\right. x \left.\right) = 6 x^{−2} .\)

\[\begin{aligned} f^{'} \left(\right. x \left.\right) & = \frac{d}{d x} \left(\right. \frac{6}{x^{2}} \left.\right) = \frac{d}{d x} \left(\right. 6 x^{−2} \left.\right) & & & \text{Rewrite} \frac{6}{x^{2}} \text{as} 6 x^{−2} . \\ & = 6 \frac{d}{d x} \left(\right. x^{−2} \left.\right) & & & \text{Apply the constant multiple rule}. \\ & = 6 \left(\right. −2 x^{−3} \left.\right) & & & \text{Use the extended power rule to differentiate} x^{−2} . \\ & = −12 x^{−3} & & & \text{Simplify}. \end{aligned}\]

Checkpoint 3.18

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

The Quotient Rule

Theorem 3.6