The Product rule and quotient rule

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Here we compute derivatives of products and quotients of functions

The product rule

Consider the product of two simple functions, say $f(x)\cdot g(x)$ where $f(x)=x^2+1$ and $g(x)=x^3-3x$ . An obvious guess for the derivative of $f(x)g(x)$ is the product of the derivatives:

$\begin{align*} f'(x)g'(x) &= (2x)(3x^2-3)\\ &= 6x^3-6x. \end{align*}$

Is this guess correct? We can check by rewriting $f$ and $g$ and doing the calculation in a way that is known to work. Write with me

$\begin{align*} f(x)g(x) &= (x^2+1)(x^3-3x)\\ &=x^5-3x^3+x^3-3x\\ &=x^5-2x^3-3x. \end{align*}$

Hence $\ddx f(x) g(x) = \ddx (x^5 - 2x^3 - 3x) = 5x^4-6x^2-3,$ so we see that $\ddx f(x) g(x) \ne f'(x)g'(x).$ So the derivative of $f(x)g(x)$ is not as simple as $f'(x)g'(x)$ . Never fear, we have a rule for exactly this situation.

The product rule If $f$ and $g$ are differentiable, then $\ddx f(x)g(x) = f(x)g'(x)+f'(x)g(x).$

Let’s return to the example with which we started.

Let $f(x)=(x^2+1)$ and $g(x)=(x^3-3x)$ . Compute: $\ddx f(x)g(x)$

Write with me $\begin{align*} \ddx f(x)g(x) &= f(x)g'(x) + f'(x)g(x)\\ &=(x^2+1)(\answer[given]{3x^2-3}) + (\answer[given]{2x})(x^3-3x). \end{align*}$

We could stop here, but we should show that expanding this out recovers our previous result. Write with me

$\begin{align*} (x^2+1)&(3x^2-3) + 2x(x^3-3x)\\ &= 3x^4-3x^2 +3x^2 -3 + 2x^4-6x^2\\ &=\answer[given]{5x^4-6x^2-3}, \end{align*}$

which is precisely what we obtained before.

Now that we are pros, let’s try one more example.

Suppose $f$ is a function whose values are given in the following table.



x	f(x)	f’(x)

1	-3	4
2	5	-1
3	2	0

Compute $\eval { \ddx \left ( (x^2+1) f(x) \right )}_{x=2}$ .

Using the product rule, write with me. $\begin{align*} \ddx\left( (x^2+1) f(x) \right) &= \ddx \left( \answer[given]{x^2+1} \right) \cdot f(x) + (x^2+1) \cdot \ddx\left( \answer[given]{f(x)} \right) \\ &= \left( \answer[given]{2x} \right) \cdot f(x) + (x^2+1) \cdot \ddx \left( f(x) \right) \end{align*}$

Plugging in $x=2$ yields:

$\begin{align*} \eval{ \ddx\left( (x^2+1) f(x) \right) }_{x=2} &= \left( \answer[given]{4} \right) \cdot f\left( \answer[given]{2} \right) + \left( \answer[given]{5} \right) \cdot f'\left( \answer[given]{2} \right)\\ &= 4 \cdot \answer[given]{5} + 5 \cdot \answer[given]{-1}\\ &= \answer[given]{15} \end{align*}$

The quotient rule

We’d like to have a formula to compute $\ddx \frac {f(x)}{g(x)}.$ This brings us to our next derivative rule.

The quotient rule If $f$ and $g$ are differentiable, then $\ddx \frac {f(x)}{g(x)} = \frac {f'(x)g(x)-f(x)g'(x)}{g(x)^2}.$

Compute: $\ddx \frac {x^2+1}{x^3-3x}$

Write with me $\begin{align*} \ddx \frac{x^2+1}{x^3-3x} &= \frac{2x(\answer[given]{x^3-3x})-(\answer[given]{x^2+1})(3x^2-3)}{(\answer[given]{x^3-3x})^2}\\ &=\frac{-x^4-6x^2+3}{(\answer[given]{x^3-3x})^2}. \end{align*}$

It is often possible to calculate derivatives in more than one way, as we have already seen. Since every quotient can be written as a product, it is always possible to use the product rule to compute the derivative, though it is not always simpler.

Compute: $\ddx \frac {625-x^2}{\sqrt {x}}$ in two ways. First using the quotient rule and then using the product rule.

First, we’ll compute the derivative using the quotient rule. Write with me $\ddx \frac {625-x^2}{\sqrt {x}} = \frac {\left (-2x\right )\left (\answer [given]{\sqrt {x}}\right ) - (\answer [given]{625-x^2})\left (\frac {1}{2}x^{-1/2}\right )}{\answer [given]{x}}.$

Second, we’ll compute the derivative using the product rule:

$\begin{align*} \ddx \frac{625-x^2}{\sqrt{x}} &= \ddx \left(625-x^2\right)x^{-1/2}\\ =\left(625-x^2\right)&\left(\answer[given]{\frac{-x^{-3/2}}{2}}\right)+ (\answer[given]{-2x})\left(x^{-1/2}\right). \end{align*}$

With a bit of algebra, both of these simplify to $-\frac {3x^2+625}{2x^{3/2}}.$

Let’s finish up with one more, mixing the rules together.

Suppose $f$ is a function whose values are given in the following table.



x	f(x)	f’(x)

1	-3	4
2	5	-1
3	2	0

Compute $\eval { \ddx \left ( \frac {x f(x)}{x+1} \right )}_{x=2}$ .

The ‘Outside Operation’ is a quotient, so we’ll start with the quotient rule. Notice that the numerator is a product? That means we’ll have to use product rule, too. $\begin{align*} \ddx\left( \frac{x f(x)}{x+1} \right) &= \frac{\ddx\left(x f(x)\right) \left(\answer[given]{x+1}\right) - \left( x f(x) \right) \ddx \left(x+1\right)}{\answer[given]{(x+1)^2}}\\ &= \frac{\left( \ddx(x) f(x) + x\ddx\left(f(x)\right) \right) (x+1) - \left( x f(x) \right) \left( \answer[given]{1}\right)}{(x+1)^2}\\ &= \frac{\left( f(x) + x f'(x) \right) (x+1) - x f(x)}{(x+1)^2} \end{align*}$

Plugging $x=2$ into this yields:

$\begin{align*} \eval{ \ddx\left( \frac{x f(x)}{x+1} \right)}_{x=2} &= \frac{\left( f\left(\answer[given]{2}\right) + x f'\left(\answer[given]{2}\right) \right) \left(\answer[given]{3}\right) - \answer[given]{2} f\left(\answer[given]{2}\right)}{\answer[given]{9}} \\ &= \frac{\left( 5 + 2 (\answer[given]{-1})\right) (3) - 2 (5)}{9} \\ &= \answer[given]{\frac{-7}{9}} \end{align*}$