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 $\frac {d}{dx} f(x) g(x) = \frac {d}{dx}(x^5 - 2x^3 - 3x) = 5x^4-6x^2-3,$ so we see that $\frac {d}{dx} f(x) g(x) \ne f'(x)g'(x).$ 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 $\frac {d}{dx} f(x)g(x) = f(x)g'(x)+f'(x)g(x).$

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

Let $f(x)=(x^2+1)$ and $g(x)=(x^3-3x)$ . Compute: $\frac {d}{dx} f(x)g(x)$

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

Using the sum rule and the product rule, write with me $\begin{align*} \frac{d}{dx} \left(xe^x-e^x \right) &=\frac{d}{dx} \left(xe^x\right)-\frac{d}{dx} e^x\\ &=(xe^x+e^x)-e^x\\ &=\answer[given]{x e^x}. \end{align*}$

The quotient rule

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

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

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

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

Compute: $\frac {d}{dx} \frac {\sin {x}}{x}$

Write with me $\begin{align*} \frac{d}{dx} \frac{\sin{x}}{x} &= \frac{\cos{x}\cdot x-\sin{x}\cdot1}{x^2} \end{align*}$

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: $\frac {d}{dx} \frac {625-x^2}{\sqrt {x}}$ 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 $\frac {d}{dx} \frac {625-x^2}{\sqrt {x}} = \frac {\left (-2x\right )\left (\answer [given]{\sqrt {x}}\right ) - (\answer [given]{625-x^2})\left (\frac {1}{2}x^{-1/2}\right )}{\answer [given]{x}}.$

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

$\begin{align*} \frac{d}{dx} \frac{625-x^2}{\sqrt{x}} &= \frac{d}{dx} \left(625-x^2\right)x^{-1/2}\\ =\left(625-x^2\right)&\left(\answer[given]{\frac{-x^{-3/2}}{2}}\right)+ (\answer[given]{-2x})\left(x^{-1/2}\right). \end{align*}$ With a bit of algebra, both of these simplify to $-\frac {3x^2+625}{2x^{3/2}}.$

Suppose we have two functions, $f$ , and $g$ , and we know that $f(4) = 3$ , $f'(4) = 5$ , $g(4) = -2$ , and $g'(4) = 2$ . What is the slope of the tangent line to the curve $y=\frac {f(x)}{g(x)}$ at the point where $x = 4$ ?

The slope of the tangent line to the curve $y=\frac {f(x)}{g(x)}$ at $x = 4$ is given by $\bigg [ \frac {d}{dx}\frac {f(x)}{g(x)} \bigg ]_{x=4}$ .

By the QuotientRule, this derivative is given by

$\begin{align*} \bigg[ \frac{d}{dx}\frac{f(x)}{g(x)} \bigg]_{x=4} &=\bigg[ \frac{f'(x)g(x)-f(x)g'(x)}{\left(g(x)\right)^2} \bigg]_{x=4}\\ &=\frac{f'(4)g(4)-f(4)g'(4)}{\left(g(4)\right)^2}\\ &=\frac{5(-2)-3\cdot2}{\left(-2\right)^2}\\ &=\frac{-10-6}{4}\\ &=4 \end{align*}$

Suppose we have two functions, $f$ , and $g$ , and we know that $f(4) = 3$ , $f'(4) = 5$ , $g(4) = -2$ , and $g'(4) = 2$ . What is the slope of the tangent line to the curve $y=\frac {xf(x)}{g(x)}$ at the point where $x = 4$ ?

The slope of the tangent line to the curve $y=\frac {xf(x)}{g(x)}$ at $x = 4$ is given by $\bigg [ \frac {d}{dx}\frac {xf(x)}{g(x)} \bigg ]_{x=4}$ .

By the QuotientRule and the Product Rule, this derivative is given by

$\begin{align*} \bigg[ \frac{d}{dx}\frac{xf(x)}{g(x)} \bigg]_{x=4} &=\bigg[ \frac{(xf'(x)+f(x))g(x)-xf(x)g'(x)}{\left(g(x)\right)^2} \bigg]_{x=4}\\ &=\frac{(4f'(4)+f(4))g(4)-4f(4)g'(4)}{\left(g(4)\right)^2}\\ &=\frac{(4\cdot5+3)(-2)-4\cdot3\cdot2}{\left(-2\right)^2}\\ &=\frac{(4\cdot5+3)(-2)-4\cdot3\cdot2}{\left(-2\right)^2}\\ &=\frac{-35}{2}\\ \end{align*}$