In the last section, we learned our first set of "shortcut" rules for differentiation, such as the Power Rule and the Sum/Difference Rules. These allow us to find derivatives much faster than using the limit definition every time. However, what happens when we have two functions that are multiplied together, like $f(x) = x^2 e^x$, or divided, like $g(x) = \frac{x^2}{x+1}$? It turns out we need two new, powerful rules to handle these situations: the Product Rule and the Quotient Rule.
It's a common and tempting mistake to think that the derivative of a product of two functions is simply the product of their derivatives. Let's see why this is incorrect.
Consider $f(x) = x$ and $g(x) = x^2$. The product is $h(x) = f(x)g(x) = x^3$. We know from the Power Rule that $h'(x) = 3x^2$. However, if we were to (incorrectly) multiply the individual derivatives, we would get $f'(x)=1$ and $g'(x)=2x$, and their product is $1 \cdot 2x = 2x$, which is not the correct answer. This shows we need a different rule.
If $f$ and $g$ are both differentiable functions, then the derivative of their product $f(x)g(x)$ is given by:
$$ \frac{d}{dx}[f(x)g(x)] = f(x) \frac{d}{dx}[g(x)] + g(x) \frac{d}{dx}[f(x)] $$In prime notation, this is written as:
$$ (fg)'(x) = f(x)g'(x) + g(x)f'(x) $$In words: "The derivative of the first function times the second, plus the first function times the derivative of the second."
Differentiate $h(x) = x^2 e^x$.
Let $f(x) = x^2$ and $g(x) = e^x$. Then we have:
Applying the Product Rule formula $(fg)' = f'g + fg'$:
$$ h'(x) = (2x)(e^x) + (x^2)(e^x) $$We can factor out $e^x$ to simplify:
$$ h'(x) = e^x(2x + x^2) $$Find the derivative of $y = (3x^2 + 1)(2x - 5)$.
Let $f(x) = 3x^2+1$ and $g(x) = 2x-5$. Then:
Applying the rule $y' = f'g + fg'$:
$$ y' = (6x)(2x - 5) + (3x^2 + 1)(2) $$Now, simplify by distributing:
$$ y' = (12x^2 - 30x) + (6x^2 + 2) $$ $$ y' = 18x^2 - 30x + 2 $$Find the derivative of $y = \sqrt{x}(x^2 + 1)$.
To understand the product rule, let's use a real-world analogy. A company's Revenue is the total amount of money it receives from sales. For example:
In general, Revenue = (Number of Items Sold) × (Price Per Item). But revenue is not static; maybe revenue goes up during holidays and down during the summer. Since both sales volume and price can change over time, we can model them as functions of time, $t$:
$$ R(t) = N(t) \cdot P(t) $$If you're the CEO, the rate at which your revenue is changing, $R'(t)$, comes from two distinct sources:
The total rate of change of revenue is the sum of these two effects:
$$ R'(t) = \underbrace{N'(t)P(t)}_{\text{Sales Effect}} + \underbrace{N(t)P'(t)}_{\text{Price Effect}} $$This shows that the product rule isn't just an arbitrary formula; it's the logical sum of all the ways a product can change when both of its factors are changing.
A company finds that the number of items it sells per month can be modeled by the function $N(t) = 100 + t^2$, where $t$ is the number of months since a new marketing campaign began. The price per item is modeled by $P(t) = 50 - t$. At what rate is the company's revenue changing at $t=5$ months?
Just as with products, we need a special rule for finding the derivative of a quotient of two functions, $Q(x) = \frac{f(x)}{g(x)}$. This rule can be derived directly from the Product Rule.
Let $Q(x) = \frac{f(x)}{g(x)}$. We want to find $Q'(x)$.
First, rewrite the equation as $f(x) = Q(x)g(x)$. Now, we can differentiate both sides using the Product Rule we just learned:
$$ f'(x) = Q'(x)g(x) + Q(x)g'(x) $$Our goal is to solve for $Q'(x)$:
$$ Q'(x)g(x) = f'(x) - Q(x)g'(x) $$ $$ Q'(x) = \frac{f'(x) - Q(x)g'(x)}{g(x)} $$Now, substitute $Q(x) = \frac{f(x)}{g(x)}$ back into the equation:
$$ Q'(x) = \frac{f'(x) - \frac{f(x)}{g(x)}g'(x)}{g(x)} $$To simplify, multiply the numerator and denominator by $g(x)$:
$$ Q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} $$And that's the Quotient Rule!
If $f$ and $g$ are both differentiable functions, then the derivative of their quotient $\frac{f(x)}{g(x)}$ is given by:
$$ \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)\frac{d}{dx}[f(x)] - f(x)\frac{d}{dx}[g(x)]}{[g(x)]^2} $$In prime notation, this is written as:
$$ \left(\frac{f}{g}\right)'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2} $$A common mnemonic is: "Low dee high, minus high dee low, square the bottom and away we go!" where "low" is $g(x)$, "high" is $f(x)$, and "dee" means "the derivative of".
Differentiate $y = \frac{x^2+x-2}{x^3+6}$.
Let $f(x) = x^2+x-2$ (high) and $g(x) = x^3+6$ (low). Then:
Applying the Quotient Rule formula $(\frac{f}{g})' = \frac{gf' - fg'}{g^2}$:
$$ y' = \frac{(x^3+6)(2x+1) - (x^2+x-2)(3x^2)}{(x^3+6)^2} $$While we could expand the numerator, it's often best to leave it in this factored form unless asked to simplify further.
Find the derivative of $F(t) = \frac{e^t}{1+e^t}$.
Let's take a moment to summarize all the rules we have learned so far. These are the building blocks for all of differential calculus.
| Rule Name | Formula |
|---|---|
| Constant Rule | $\frac{d}{dx}(c) = 0$ |
| Power Rule | $\frac{d}{dx}(x^n) = nx^{n-1}$ |
| Constant Multiple Rule | $\frac{d}{dx}[cf(x)] = c \frac{d}{dx}f(x)$ |
| Sum/Difference Rule | $\frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}f(x) \pm \frac{d}{dx}g(x)$ |
| Natural Exponential Rule | $\frac{d}{dx}(e^x) = e^x$ |
| Product Rule | $(fg)' = f'g + fg'$ |
| Quotient Rule | $\left(\frac{f}{g}\right)' = \frac{gf' - fg'}{g^2}$ |
Find the derivative of $f(x) = (1+2x)e^x$.
Find the equation of the tangent line to the curve $y = \frac{2x}{x+1}$ at the point $(1,1)$.
Differentiate $g(x) = \frac{1}{x}(x^2 + e^x)$.
Today we added two essential tools, the Product Rule and the Quotient Rule, to our calculus toolkit. These rules allow us to differentiate much more complex functions that are formed by multiplying or dividing simpler functions. By combining these with our previous rules, we can now find the derivative of a vast array of functions. In our next lecture, we will continue to expand our capabilities by learning how to differentiate trigonometric functions.