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.

Compute: $\ddx ((x^3+2x^2)(\sqrt {x}+2))$

Using the product rule and the sum rule, write with me $\ddx \left ((x^3+2x^2)(\sqrt {x}+2) \right ) =\answer [given]{(3 x^2 + 4x)(\sqrt {x} + 2) + (x^3 + 2 x^2)(1/(2 \sqrt {x}))}.$

The quotient rule

We’d like to have a formula to compute $\ddx \frac {f(x)}{g(x)}$ if we already know $f'(x)$ and $g'(x)$ . Instead of attacking this problem head-on, let’s notice that we’ve already done part of the problem: $f(x)/g(x)= f(x)\cdot (1/g(x))$ , that is, this is really a product, and we can compute the derivative if we know $f'(x)$ and $(1/g(x))'$ . 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 $-\frac {3x^2+625}{2x^{3/2}}.$