Combining differentiation rules
A function such as x sin(x^2) divided by e^x is built with a product, a quotient and a function inside a function all at once. Which rule do you apply first, and how do you keep track of the rest?
Where you are 13 in Derivatives
When you reach for it
Each rule so far answers one shape: a power, an exponential, a sine, a product, a quotient, a composition. Real functions use several shapes at once. In
there is a quotient, a product inside its numerator and a composition inside that product. No new rule is needed. What is needed is a way to see which rule applies first, to pass the smaller pieces on to the other rules in turn, and to pick the form of the function that makes the work shortest. Reach for this whenever a function is built from more than one operation, which from now on is almost always.
The lesson runs over two days. Day 1 reads the structure of an expression, peels its layers, chooses between the quotient rule and a rewrite, simplifies, and checks an answer with numbers. Day 2 goes deeper: several rules nested inside each other, functions known only through a table or a graph, and the choice of least work.
The procedure
Day 1: reading the structure
The rules available. With f and g differentiable:
| Shape | Derivative |
|---|---|
| x^n | nx^{n-1} |
| e^x | e^x |
| a^x | a^x \ln a |
| \sin x | \cos x |
| \cos x | -\sin x |
| \tan x | \sec^2 x |
| \sec x | \sec x\tan x |
| \csc x | -\csc x\cot x |
| \cot x | -\csc^2 x |
| f(x)g(x) | f'(x)g(x) + f(x)g'(x) |
| \dfrac{f(x)}{g(x)} | \dfrac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} |
| f\big(g(x)\big) | f'\big(g(x)\big)\,g'(x) |
The power rule is in The power rule and basic differentiation rules, the exponentials in Derivatives of exponential functions, and all six trigonometric functions in Derivatives of the trigonometric functions. Sums and constant multiples are differentiated term by term throughout.
The outermost operation decides the first rule. Imagine computing the function at a number, say x = 2, by hand. Whatever you do last is the top level of the expression. For \sin^4 x you find \sin 2 and then raise it to the fourth power, so the top level is a power of a function: the chain rule with the power rule outside. For \sin(x^4) the last step is the sine, so the chain rule has sine outside. For x^4\sin x the last step is a multiplication, and for \displaystyle \frac{\sin x}{x^4} a division. The derivative is taken in the reverse of the order of evaluation: the last operation done is the first rule applied.
Each rule hands back smaller problems. The product rule asks for f' and g'; the chain rule asks for the derivative of the outer function, evaluated at the inside, and for the derivative of the inside. Each of those is a new, smaller expression with its own outermost operation, and the same reading applies to it. The work stops when every piece is in the table. Leibniz notation keeps the bookkeeping honest: name each layer, and the derivative is the product of one factor per link, as in Figure 1.
A quotient need not use the quotient rule. When the numerator is a constant, \displaystyle \frac{c}{g(x)} = c\,g(x)^{-1}, and the chain rule is shorter. When the denominator is a single power of x, dividing it into each term of the numerator leaves a sum of powers. When the denominator is a power of a function, writing f(x)\,g(x)^{-n} and using the product and chain rules gives the same answer as the quotient rule, and the choice is a matter of which algebra is shorter. Both are correct, because \displaystyle \frac{f}{g} = f \cdot g^{-1} wherever g \neq 0; the quotient rule was proved from exactly that identity.
Simplify before and after. An identity or a cancellation before differentiating can shrink the work a great deal, but the simpler expression equals the original only where the original is defined, and the derivative carries that restriction. After differentiating, factoring out common factors turns a long answer into one whose zeros and sign can be read off.
Check with numbers. For a small h, the symmetric difference quotient
is the slope of a secant line straddling a. It is the average of the right-hand and left-hand difference quotients from the definition, so when f is differentiable at a it approaches f'(a) as h \to 0. Two or three agreeing digits are strong evidence that a formula is right; a disagreement in the first digit shows a slip somewhere.
Day 2: deeper layers, tables and graphs
Deeper nesting is the same reading, repeated. A product can sit inside a composition, and a quotient inside a product. Applying the product rule twice gives the rule for three factors:
one term for each factor taking its turn to be differentiated.
Tables and graphs. The rules are statements about numbers at a point, so they work when f and g are known only through values. To find (f \circ g)'(a) you need g(a), g'(a), and f' at the number g(a), which is usually a different row of a table. On a graph made of straight segments, a slope is read off each segment, and at a corner it does not exist.
Least work. Before starting, ask which form of the function gives the shortest correct path: a quotient written as a product, a power brought up out of a denominator, a factor canceled, an identity used. The answer is the same function whichever path is taken; only the length of the work changes.
Worked examples
Day 1: the outermost operation
Differentiate \sin^4 x, \sin(x^4), x^4\sin x and \dfrac{\sin x}{x^4}.
SolutionThe four use the same two pieces, \sin x and x^4, joined in four different ways, so each needs a different first rule.
\sin^4 x means (\sin x)^4: the last operation is the fourth power. Chain rule, power outside, sine inside:
\sin(x^4) ends with the sine. Chain rule, sine outside, x^4 inside:
x^4\sin x ends with a multiplication. Product rule:
\dfrac{\sin x}{x^4} ends with a division. The denominator is a single power of x, but it cannot be divided into \sin x term by term, so use the quotient rule (or, equally, \sin x \cdot x^{-4} with the product rule):
after canceling a common factor x^3 from top and bottom, for x \neq 0.
Day 1: peeling layers, with a numerical check
Find \dfrac{dy}{dx} for \displaystyle y = \sqrt{1 + \sin^2(2x)} and its value at \displaystyle x = \frac{\pi}{8}. Then Tiina checks the value with a symmetric difference quotient, h = 0.001.
SolutionEvaluating at a number, the last step is the square root, so the chain rule with \displaystyle \sqrt{u} outside comes first. Inside is u = 1 + w^2 with w = \sin v and v = 2x. Figure 1 shows the layers; the chain rule multiplies one factor per link:
Putting the layers back in terms of x:
At \displaystyle x = \frac{\pi}{8}, \displaystyle 2x = \frac{\pi}{4} and \displaystyle \sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}, so the numerator is \displaystyle 2 \cdot \frac{1}{2} = 1 and the denominator is \displaystyle \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}:
Tiina’s check: with \displaystyle a = \frac{\pi}{8} \approx 0.392699, a calculator in radian mode gives y(a + 0.001) \approx 1.2255611 and y(a - 0.001) \approx 1.2239281, so
The two agree to five digits.
Day 1: quotient rule or rewrite
Differentiate \displaystyle r(x) = \dfrac{4}{\sqrt{x^2 + 9}} and \displaystyle s(x) = \dfrac{x^2}{\sqrt{x^2 + 9}}, and find r'(4) and s'(4).
SolutionThe numerator of r is a constant, so the quotient rule would spend half its effort differentiating 4. Write r(x) = 4(x^2 + 9)^{-1/2}; the top level is a constant times a power of a function, and the chain rule gives
At x = 4, x^2 + 9 = 25 and 25^{3/2} = 125, so \displaystyle r'(4) = -\frac{16}{125}.
For s the numerator varies. Write s(x) = x^2(x^2 + 9)^{-1/2} and use the product rule, with the chain rule on the second factor:
Factor out the lower power (x^2 + 9)^{-3/2}; what remains of the first term is 2x(x^2 + 9):
The quotient rule gives
and multiplying top and bottom by \displaystyle \sqrt{x^2 + 9} leads to the same result. Then \displaystyle s'(4) = \frac{64 + 72}{125} = \frac{136}{125}.
Day 1: simplify before, factor after
(a) Differentiate h(x) = \dfrac{e^{2x} - 1}{e^x - 1}.
(b) Differentiate k(x) = (x^2 + 1)^3(2x - 1)^2 and find every point where the tangent line is horizontal.
Solution(a) The quotient rule works but is long. Since e^{2x} = (e^x)^2, the numerator is a difference of squares, (e^x - 1)(e^x + 1). Where e^x \neq 1, that is for x \neq 0, the common factor cancels:
At x = 0 the original quotient is \displaystyle \frac{0}{0}, so h is not defined there and neither is h'(0), even though e^x is.
(b) The top level is a product; each factor is a power of a function. Product rule, with the chain rule twice:
Both terms contain (x^2 + 1)^2 and (2x - 1), and both have a factor 2. Factor them out:
Now the zeros can be read off. x^2 + 1 \geq 1 is never 0; 8x^2 - 3x + 2 has discriminant 9 - 64 < 0, so it is never 0 either. Only 2x - 1 = 0 remains: the tangent is horizontal at \displaystyle x = \frac{1}{2} alone, at the point \displaystyle \left(\frac{1}{2}, 0\right) since \displaystyle k\left(\frac{1}{2}\right) = 0. The unfactored form hides this completely.
Day 2: deeper layers and the least work
(a) Jaanus differentiates y = \dfrac{x\sin(x^2)}{e^{x}}. Find y' and y'(1).
(b) Differentiate g(x) = \sin\!\left(\dfrac{\pi x}{x^2 + 1}\right) and find g'(0) and g'(1).
Solution(a) The top level is a quotient. Jaanus notices that dividing by e^x is multiplying by e^{-x}, so y = x \cdot \sin(x^2) \cdot e^{-x}, a product of three factors with no quotient left. Their derivatives are 1, 2x\cos(x^2) by the chain rule, and -e^{-x} by the chain rule. The three-factor product rule:
Every term has e^{-x}; factoring it out and collecting the sine terms,
At x = 1 the sine term vanishes: \displaystyle y'(1) = \frac{2\cos 1}{e} \approx 0.3975. Mihkel uses the quotient rule instead, with the product rule for the numerator; his answer has e^{2x} in the denominator and an e^x in every term of the numerator, and after canceling it is the same expression. The rewrite saved one rule and one cancellation.
(b) The last operation is the sine, so the chain rule comes first, and the inside is a quotient. By the quotient rule,
so
At x = 0: \cos 0 = 1 and the fraction is \pi, so g'(0) = \pi. At x = 1 both factors are 0, since the inside is \displaystyle \frac{\pi}{2} and 1 - x^2 = 0; so g'(1) = 0.
Day 2: values from a table
The table gives values of differentiable functions f and g.
| x | f(x) | f'(x) | g(x) | g'(x) |
|---|---|---|---|---|
| 1 | 3 | -1 | 2 | 4 |
| 2 | 1 | 5 | 4 | -2 |
| 4 | 2 | 3 | 1 | 6 |
Find p'(2) for p(x) = x\,f(x^2), q'(1) for q(x) = \dfrac{g(x)}{f(2x)}, and r'(2) for r(x) = e^{f(g(x))}.
Solutionp is a product whose second factor is a composition. Product rule, then the chain rule on f(x^2):
At x = 2 the inside is 4, so the row x = 4 is needed: p'(2) = f(4) + 2 \cdot f'(4) \cdot 4 = 2 + 2 \cdot 3 \cdot 4 = 26.
q is a quotient with a composition in the denominator; the derivative of f(2x) is 2f'(2x). Quotient rule:
At x = 1 this uses g(1) = 2, g'(1) = 4, f(2) = 1, f'(2) = 5: \displaystyle q'(1) = \frac{4 \cdot 1 - 2 \cdot 2 \cdot 5}{1^2} = -16.
r has three layers: the exponential, then f, then g. One factor per layer:
At x = 2: g(2) = 4, then f(4) = 2 and f'(4) = 3, and g'(2) = -2. So r'(2) = e^2 \cdot 3 \cdot (-2) = -6e^2 \approx -44.33.
Day 2: slopes read from a graph
Figure 2 shows f. Find u'(1) for u(x) = x^2 f(x) and v'(3) for v(x) = \dfrac{f(x)}{e^{x}}, and decide whether w'(1) exists for w(x) = f(x^2 + 1).
From the graph, f has slope 1 on (0, 2), slope -1 on (2, 5), and slope 0 on (5, 6). So f(1) = 3 and f'(1) = 1; f(3) = 3 and f'(3) = -1.
u is a product: u'(x) = 2x f(x) + x^2 f'(x), so u'(1) = 2 \cdot 3 + 1 \cdot 1 = 7.
For v, write f(x)e^{-x}; the product rule gives v'(x) = \big(f'(x) - f(x)\big)e^{-x}, so \displaystyle v'(3) = \frac{-1 - 3}{e^3} = -\frac{4}{e^3} \approx -0.199.
For w the chain rule would give f'(x^2 + 1) \cdot 2x, but at x = 1 the inside is 2, a corner of f, where f'(2) does not exist. The chain rule’s hypothesis fails, so look directly. For x a little below 1 the inside is a little below 2, where f has slope 1, and w has slope 1 \cdot 2x \to 2. For x a little above 1 the inside is above 2, slope -1, and w has slope -1 \cdot 2x \to -2. The one-sided slopes are 2 and -2, so w'(1) does not exist: w has a corner at x = 1.
Boundaries
Each rule needs its pieces differentiable at the right place. The product and quotient rules need f and g differentiable at a (and g(a) \neq 0 for the quotient); the chain rule needs the inside differentiable at a and the outside differentiable at the number g(a), which is where Example 7 broke down.
A failed hypothesis means only that the rule is silent, not that the derivative is missing. The chain rule cannot be applied to \displaystyle \sqrt{x^4} at x = 0, because \displaystyle \sqrt{u} has no derivative at u = 0, yet \displaystyle \sqrt{x^4} = x^2 has derivative 0 there. When a rule does not apply, return to one-sided slopes or to the definition.
A simplification holds only on the domain of the original function. In Example 4(a), h(x) = e^x + 1 for x \neq 0 only, and h'(0) does not exist.
A numerical check is evidence, not proof. An h that is too small loses digits to rounding, and at a corner the symmetric quotient averages the two one-sided slopes: for w in Example 7 it gives a value near 0 at x = 1, though no derivative exists there.
Where this goes next
With every rule combinable, the remaining derivatives of this unit come quickly. The derivative of the natural logarithm and the inverse trigonometric functions add new rows to the table, and implicit differentiation applies the chain rule to every term of an equation in x and y. Combined rules are also what make second derivatives substantial, since the derivative of a product is a sum of products, and they carry every later application: tangent lines, related rates, and the search for maxima and minima.
Exercises
Day 1: the outermost operation
Name the outermost operation, then differentiate.
- y = x^3 \tan x
- y = \tan(x^3)
- y = \tan^3 x
- y = \dfrac{\tan x}{x^3}
Day 1: peeling layers
- y = \cos^2(4x + 1)
- \displaystyle y = e^{\sqrt{x^2 + 3}}; also find y'(1).
- \displaystyle y = \sqrt{2 + \cos(\pi x)}; also find \displaystyle y'\big(\frac{1}{2}\big).
Day 1: quotient rule or rewrite
Choose the shortest route.
- \displaystyle y = \dfrac{3}{\sqrt{1 + e^{x}}}
- \displaystyle y = \dfrac{x}{\sqrt{x + 1}}
- y = \dfrac{e^{3x} + e^{x}}{e^{x}}
Day 1: simplify and check
- Simplify y = \dfrac{1 - \cos^2 x}{\sin x} before differentiating, and say where the result holds.
- Differentiate y = x^2(x - 3)^3, factor the result completely, and find every x where the tangent is horizontal.
- Mihkel finds \displaystyle \dfrac{d}{dx}\big[\sqrt{x}\cos x\big] = \dfrac{\cos x}{2\sqrt{x}} - \sqrt{x}\sin x. Check his formula at x = 1 against a symmetric difference quotient with h = 0.01.
Day 2: deeper layers
- y = x\,e^{\cos x}
- y = \dfrac{e^{2x}}{1 + \sin x}
- y = \cos\!\big(x^2 e^{x}\big)
- y = \left(\dfrac{x}{x + 2}\right)^{4}; simplify fully.
- \displaystyle y = \dfrac{x^2 e^{x}}{\sqrt{x}}, by the least work you can find.
Day 2: tables and graphs
Exercises 19 to 21 use this table of differentiable functions f and g.
| x | f(x) | f'(x) | g(x) | g'(x) |
|---|---|---|---|---|
| 0 | 2 | 3 | 1 | -1 |
| 1 | 0 | 4 | 3 | 2 |
| 3 | 1 | 2 | 0 | 5 |
- h(x) = f(x)\,e^{g(x)}; find h'(0).
- k(x) = \dfrac{f(x)}{1 + g(x)^2}; find k'(1).
- \displaystyle m(x) = \sqrt{g(f(x))}; find m'(3).
Exercises 22 and 23 use the function f of Figure 2.
- p(x) = \dfrac{x}{f(x)}; find p'(4).
- q(x) = f(x)^3; find q'(4), and decide whether q'(5) exists.
Reasoning
- Jaanus differentiates \dfrac{\tan x}{\sec x} with the quotient rule; Tiina simplifies first. Carry out both and show they agree. Where does the result hold?
- Let u be differentiable with u(x) > 0. Explain why \displaystyle \sqrt{u(x)} has a horizontal tangent exactly where u'(x) = 0, and use this to find where \displaystyle y = \sqrt{x^2 - 6x + 10} has one.
Answers
- Product: 3x^2\tan x + x^3\sec^2 x.
- Composition, tangent outside: 3x^2\sec^2(x^3).
- Power of a function: 3\tan^2 x\sec^2 x.
- Quotient: \dfrac{x^3\sec^2 x - 3x^2\tan x}{x^6} = \dfrac{x\sec^2 x - 3\tan x}{x^4}, canceling x^2.
- -8\cos(4x + 1)\sin(4x + 1). Power outside: 2\cos(4x + 1), times -\sin(4x + 1), times 4.
- \displaystyle y' = \dfrac{x\,e^{\sqrt{x^2 + 3}}}{\sqrt{x^2 + 3}}; y'(1) = \dfrac{e^2}{2}. The derivative of \displaystyle \sqrt{x^2 + 3} is \displaystyle \frac{2x}{2\sqrt{x^2 + 3}}; at x = 1 the root is 2.
- \displaystyle y' = -\dfrac{\pi\sin(\pi x)}{2\sqrt{2 + \cos(\pi x)}}; \displaystyle y'\big(\frac{1}{2}\big) = -\dfrac{\pi}{2\sqrt{2}} = -\dfrac{\pi\sqrt{2}}{4} \approx -1.111, since \displaystyle \sin\frac{\pi}{2} = 1 and \displaystyle \cos\frac{\pi}{2} = 0.
- -\dfrac{3e^x}{2(1 + e^x)^{3/2}}. Write 3(1 + e^x)^{-1/2}; chain rule: \displaystyle 3 \cdot \left(-\frac{1}{2}\right)(1 + e^x)^{-3/2} \cdot e^x.
- \dfrac{x + 2}{2(x + 1)^{3/2}}. As x(x + 1)^{-1/2}: \displaystyle (x + 1)^{-1/2} - \frac{x}{2}(x + 1)^{-3/2}; factor out \displaystyle \frac{1}{2}(x + 1)^{-3/2} to leave 2(x + 1) - x.
- 2e^{2x}. Divide first: y = e^{2x} + 1.
- y = \sin x and y' = \cos x, wherever \sin x \neq 0, that is x \neq k\pi for every integer k. The numerator is \sin^2 x by the Pythagorean identity.
- y' = x(x - 3)^2(5x - 6); horizontal tangents at x = 0, \displaystyle x = \frac{6}{5}, x = 3. Product rule gives 2x(x - 3)^3 + 3x^2(x - 3)^2; factor out x(x - 3)^2 to leave 2(x - 3) + 3x = 5x - 6.
- They agree: the formula gives \displaystyle \frac{\cos 1}{2} - \sin 1 \approx -0.57132, and with f(1.01) \approx 0.534513 and f(0.99) \approx 0.545940 the quotient is \dfrac{0.534513 - 0.545940}{0.02} \approx -0.57131.
- e^{\cos x}(1 - x\sin x). Product rule; the derivative of e^{\cos x} is -\sin x\,e^{\cos x}.
- \dfrac{e^{2x}(2 + 2\sin x - \cos x)}{(1 + \sin x)^2}. Quotient rule: \displaystyle \frac{2e^{2x}(1 + \sin x) - e^{2x}\cos x}{(1 + \sin x)^2}.
- -(2x + x^2)e^x\sin\!\big(x^2 e^x\big). Chain rule outside, product rule for the inside: (x^2e^x)' = 2xe^x + x^2e^x.
- \dfrac{8x^3}{(x + 2)^5}. Chain rule: \displaystyle 4\left(\frac{x}{x + 2}\right)^3 times \displaystyle \frac{(x + 2) - x}{(x + 2)^2} = \frac{2}{(x + 2)^2}.
- \left(\tfrac{3}{2}x^{1/2} + x^{3/2}\right)e^x, for x > 0. Simplify first: y = x^{3/2}e^x, then the product rule.
- h'(0) = e. h' = f'e^{g} + f\,e^{g}g', so 3e + 2e(-1) = e.
- k'(1) = \dfrac{2}{5}. \displaystyle k' = \frac{f'(1 + g^2) - f \cdot 2gg'}{(1 + g^2)^2} = \frac{4 \cdot 10 - 0}{100}.
- \displaystyle m'(3) = \dfrac{2\sqrt{3}}{3}. \displaystyle m' = \frac{g'(f(x))\,f'(x)}{2\sqrt{g(f(x))}}; f(3) = 1, g(1) = 3, g'(1) = 2, f'(3) = 2, so \displaystyle \frac{4}{2\sqrt{3}}.
- p'(4) = \dfrac{3}{2}. f(4) = 2, f'(4) = -1; \displaystyle p' = \frac{f(x) - xf'(x)}{f(x)^2} = \frac{2 + 4}{4}.
- q'(4) = 3f(4)^2f'(4) = 3 \cdot 4 \cdot (-1) = -12. q'(5) does not exist: f(5) = 1, and the one-sided slopes of q at 5 are 3 \cdot 1 \cdot (-1) = -3 and 3 \cdot 1 \cdot 0 = 0.
- Both give \cos x, wherever \cos x \neq 0. Tiina: \displaystyle \frac{\tan x}{\sec x} = \frac{\sin x}{\cos x} \cdot \cos x = \sin x. Jaanus, using \sec^2 x - \tan^2 x = 1 in the last step:
Neither \tan x nor \sec x is defined where \cos x = 0. 25. At x = 3. By the chain rule \displaystyle \frac{d}{dx}\sqrt{u} = \frac{u'(x)}{2\sqrt{u(x)}}, and the denominator is a positive number, so the fraction is 0 exactly when u'(x) = 0. Here u' = 2x - 6, which is 0 at x = 3, and u(3) = 1 > 0.