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The five rules, where each comes from, and how to read an expression to know which one applies.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to differentiate a polynomial term by term, apply the product and quotient rules with the two terms and the order in the right places, rewrite a negative or fractional power so that the power rule does the work a quotient rule would otherwise do, choose a rule by reading the outermost operation of an expression, compose the rules where one sits inside another, and apply them to functions given only as values in a table.
Differentiating from the definition works and is slow. These five rules do it instead, and each is proved once from the definition so that it never has to be used again.
| Rule | Statement |
|---|---|
| constant | $(c)' = 0$ |
| power | $(x^n)' = nx^{n-1}$, for every real $n$ |
| constant multiple | $(cf)' = cf'$ |
| sum | $(f + g)' = f' + g'$ |
| product | $(fg)' = f'g + fg'$ |
| quotient | $\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$, where $g \ne 0$ |
The first four are unsurprising. The last two are the ones worth stating carefully, because the natural guesses are both wrong:
$(fg)' \ne f'g'$. The product rule has two terms because a product changes for two reasons — $f$ moved, and $g$ moved — and both contributions have to be counted. The proof makes this visible: add and subtract $f(x+h)g(x)$ in the numerator, and the two pieces that fall out are exactly the two terms.
$\left(\dfrac{f}{g}\right)' \ne \dfrac{f'}{g'}$. And the order in the quotient rule's numerator is not free: $f'g - fg'$, numerator's derivative first. Swapping it negates the answer, which is the commonest error in this lesson and one that no amount of care with the algebra will catch, because the algebra is correct either way.
Choosing a rule is reading structure. Ask what the last operation is — the one performed after everything else — and that names the rule. Everything inside is handled by a further application, which is what makes the rules compose.
Another way: picture
A rectangle of width $f$ and height $g$, so its area is the product. Increase $x$ a little: the width grows by $f'\,dx$ and the height by $g'\,dx$, adding a strip along the top of area $f \cdot g'\,dx$, a strip along the side of area $f' \, dx \cdot g$, and a tiny corner square of area $f'g'\,(dx)^2$. The two strips are the two terms of the product rule; the corner is negligible compared with $dx$, which is exactly what the limit discards.
Another way: steps
The trick is one line, and it is the same trick that proves the limit laws: add and subtract a convenient middle term.
$$\frac{f(x+h)g(x+h) - f(x)g(x)}{h} = \frac{f(x+h)g(x+h) - f(x+h)g(x)}{h} + \frac{f(x+h)g(x) - f(x)g(x)}{h}$$ $$= f(x+h)\,\frac{g(x+h) - g(x)}{h} + g(x)\,\frac{f(x+h) - f(x)}{h}.$$
As $h \to 0$: the second factor of the first term tends to $g'(x)$; the first factor tends to $f(x)$, because $f$ is differentiable and therefore continuous — which is where the previous lesson's implication gets used. The second term tends to $g(x)f'(x)$.
So $(fg)' = fg' + gf'$. The quotient rule then follows from the product rule applied to $f = (f/g) \cdot g$, rearranged; it is not an independent fact.
$(fg)' = f'g'$. The most common wrong rule in the subject. Test it on $f = g = x$: the product is $x^2$ with derivative $2x$, while $f'g' = 1$.
Reversing the quotient rule's numerator. $fg' - f'g$ over $g^2$ is the negative of the right answer, and nothing downstream will flag it.
Using the quotient rule on a constant numerator. $\dfrac{5}{x^3}$ is $5x^{-3}$; the power rule is one step and the quotient rule is four.
Forgetting that the power rule covers all real exponents. $x^{1/2}$, $x^{-3}$ and $x^{\pi}$ all obey it. What it does not cover is a variable exponent: $2^x$ is not $x2^{x-1}$, because the variable is in the wrong place.
Differentiating a product by expanding, when the factors are not polynomials. Expanding works when it works; the product rule always does, which is why it is worth having even where expansion is available.
The rules are often learned as shapes to match, and then they get applied to shapes they superficially resemble. $(f \circ g)'$ is not $f' \circ g'$; $(f^g)'$ is not anything on this list; $\left(\dfrac{1}{f}\right)'$ is not $\dfrac{1}{f'}$.
The protection is to remember what each rule says. The product rule says a product changes for two reasons and both must be counted. Once that is the sentence in your head rather than $u'v + uv'$, the wrong versions stop being tempting: $f'g'$ counts neither reason, and it is obvious that it counts neither.
$f(x) = \dfrac{x^3 + 2x}{x}$. The quotient rule applies and there is no need for it.
Look before leaping.
Divide term by term: $f(x) = x^2 + 2$ for $x \ne 0$.
Now it is a polynomial.
$f'(x) = 2x$. The quotient rule gives the same answer after a good deal more work — and note the domain: the original is undefined at $0$, so $f'$ is too.
$f(x) = \dfrac{x^2(x+1)}{x - 3}$. The outermost operation is the division, so the quotient rule is the frame.
Structure first.
$u = x^2(x+1)$ and $v = x - 3$. Differentiating $u$ needs the product rule inside: $u' = 2x(x+1) + x^2 = 3x^2 + 2x$.
Rules compose; one is inside the other.
$f' = \dfrac{(3x^2+2x)(x-3) - x^2(x+1)}{(x-3)^2}$, which can be expanded or left as it is depending on what is wanted next.
Simplify only when there is a reason.
Product rule with $u = 2x+1$, $v = x^2 - 3$: $u' = 2$, $v' = 2x$.
Write all four down.
$f' = 2(x^2 - 3) + (2x+1)(2x) = 2x^2 - 6 + 4x^2 + 2x$.
$= 6x^2 + 2x - 6$. Check by expanding first: $f = 2x^3 + x^2 - 6x - 3$, and the power rule agrees.
Match each common derivative error to the missing rule feature.
| Missing the two product-rule terms | Missing the squared denominator | Dropped the constant multiplier | Confused a sum with a product | |
|---|---|---|---|---|
| $(uv)'=u'v'$ | ||||
| $(u/v)'=(u'v-uv')/v$ | ||||
| $(7f)'=f'$ | ||||
| $(f+g)'=fg$ |
Match each expression shape to the rule that must appear.
| Use the sum rule term by term | Use the product rule | Use the quotient rule or rewrite first | Use the chain rule | |
|---|---|---|---|---|
| $x^3+\sin x$ | ||||
| x^2e^x | ||||
| (x+1)/(x-1)$ | ||||
| (3x-1)^5$ |
Match each expression's outermost structure to the first differentiation rule to apply.
| Linearity and the power rule term by term | The product rule | The quotient rule | The chain rule after identifying the outer power | |
|---|---|---|---|---|
| $3x^4-2x+7$ | ||||
| $(x^2+1)(x-4)$ | ||||
| $(x+1)/(x^2+1)$ | ||||
| $(2x-3)^5$ |
Differentiate $f(x) = 2x^3 + 5x^2 + 9x + 3$.
Answer:
For $f(x) = 4x^2(x + 4)$, write $f'(x)$ using the product rule.
Answer:
At $x = 3$: $u = 9$, $u' = 7$, $v = 2$, $v' = 3$. Find $(uv)'$ at $3$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $f(x) = \dfrac{3x}{x + 4}$, find $f'(3)$.
Answer:
You can differentiate with all five rules and pick the right one by structure. Say in your own words why the product rule has two terms. Next: the chain rule, which is what makes the rules reach beyond the expressions you can take apart.
8. Your turn: differentiate $f(x) = (2x + 1)(x^2 - 3)$, step 3
Two routes, one answer.