5.6 Chain, product and quotient rules

Polynomials are only a small part of the functions used to model the world. Oscillations are described by sine and cosine, growth and decay by $e^x$, and many scientific scales by $\ln x$. Real models also build new functions out of old ones: a vibration that dies away is a product such as $e^{-t}\sin t$, the height of a capsule on a rotating wheel is a cosine of a multiple of time, and an average is a quotient.

Differentiation of these functions rests on a short list of standard derivatives and three rules for combining them. The chain rule handles a function of a function, the product rule a product, and the quotient rule a quotient. Together they allow the derivative of almost any function met in the course to be found, so that its gradient, tangents and rates of change can be studied.

Derivatives of standard functions

Rational powers of x

The power rule $\frac{d}{dx}\left(x^n\right) = nx^{n-1}$ was met in 5.3 for integer powers. It holds for every real power $n$, so in particular for every rational power, such as $\frac{1}{2}$, $\frac{3}{2}$ or $-\frac{1}{3}$. This means that roots can be differentiated once they are written as powers:

The result of differentiating $x^n$ is in the formula booklet:

The derivatives of sin x and cos x

The gradient of the graph of $y = \sin x$ can be read from the curve. At $x = 0$ the curve rises at its steepest, at $x = \frac{\pi}{2}$ it has a maximum and gradient $0$, at $x = \pi$ it falls at its steepest, and at $x = \frac{3\pi}{2}$ it has a minimum. Plotting these gradients against $x$ gives a curve that starts at $1$, falls to $0$, reaches $-1$ and returns: the graph of $y = \cos x$.

The same reasoning applied to $y = \cos x$ produces a curve that starts at $0$, falls to $-1$ at $\frac{\pi}{2}$ and rises again: the graph of $y = -\sin x$. Both results are in the formula booklet:

The graphs show only that these results are reasonable. They can be proved from first principles, using the limit $\frac{\sin h}{h} \to 1$ as $h \to 0$, which is true only when angles are measured in radians.

The derivative of e to the x

The exponential function $e^x$ (see 2.9) has a remarkable property: its derivative is itself.

So the gradient of the graph of $y = e^x$ at any point equals the $y$-coordinate of that point. At $(0, 1)$ the gradient is $1$, at $(1, e)$ it is $e$, and at $(2, e^2)$ it is $e^2 \approx 7.39$. This property is what defines the number $e$: among all exponential functions $a^x$, only $a = e$ gives a graph whose gradient at $(0, 1)$ is exactly $1$. For other bases, the gradient is proportional to the value, with a constant different from 1, which 5.15 shows is $\ln a$.

The derivative of ln x

The natural logarithm $\ln x$ is the inverse of $e^x$, so its graph is the reflection of $y = e^x$ in the line $y = x$. Reflection in $y = x$ swaps the rise and the run of every tangent, so it replaces each gradient by its reciprocal. The point $(\ln a, a)$ on $y = e^x$, where the gradient is $a$, becomes the point $(a, \ln a)$ on $y = \ln x$, where the gradient must be $\frac{1}{a}$. This gives the result in the formula booklet:

The graph of $y = \ln x$ always increases, since $\frac{1}{x} > 0$ for $x > 0$, but it becomes less and less steep as $x$ grows. This matches the slow growth of a logarithmic scale.

Standard derivatives

All five results are in the formula booklet. As with polynomials, the derivative of a sum is the sum of the derivatives, and a constant multiple stays as a multiple: $\frac{d}{dx}\big(af(x) + bg(x)\big) = af'(x) + bg'(x)$.

The laws of logarithms and exponents (see 1.7) can turn a function that looks complicated into a sum of standard ones. For example, $\ln(5x) = \ln 5 + \ln x$, so $\frac{d}{dx}\ln(5x) = \frac{1}{x}$: the constant $\ln 5$ differentiates to $0$. Likewise $\ln\left(x^3\right) = 3\ln x$ has derivative $\frac{3}{x}$, and $e^{x + 2} = e^2 \times e^x$ has derivative $e^2 e^x = e^{x+2}$.

The chain rule

Composite functions

A composite function applies one function to the output of another (see 2.5). In $y = (3x^2 - 5)^4$, the input $x$ is first turned into $3x^2 - 5$, and then this result is raised to the power 4. The first step is the inner function, $u = 3x^2 - 5$, and the second is the outer function, $y = u^4$.

Expanding $(3x^2 - 5)^4$ and differentiating term by term would work, but would take a long time, and for $\sin(x^2)$ or $e^{-2x}$ there is nothing to expand. The chain rule differentiates a composite function directly.

Why rates multiply

Suppose $u$ changes 3 times as fast as $x$, and $y$ changes 2 times as fast as $u$. Then a small change in $x$ produces a change 3 times as large in $u$, which in turn produces a change 2 times as large again in $y$. So $y$ changes $2 \times 3 = 6$ times as fast as $x$. Rates of change along a chain of functions multiply.

In terms of small changes $\delta x$, $\delta u$ and $\delta y$, the fractions cancel:

As $\delta x \to 0$, $\delta u \to 0$ too, and each ratio tends to the corresponding derivative. This gives the chain rule, in the form printed in the formula booklet.

The formula looks as if the $du$ terms cancel. They do not cancel literally, since $\frac{dy}{du}$ is not a fraction, but the notation was designed so that the rule is easy to remember and use correctly.

Using the chain rule with a substitution

When first learning the rule, it helps to name the inner function $u$ and to write out each derivative.

With practice, the substitution can be done mentally: "differentiate the bracket as if it were $x$, then multiply by the derivative of what is inside the bracket". The same pattern gives general results for each standard outer function.

When the inner function is linear, $g(x) = ax + b$, its derivative is the constant $a$. The derivative is then just the derivative of the outer function multiplied by $a$: $\frac{d}{dx}e^{3x} = 3e^{3x}$, $\frac{d}{dx}\cos(2x - 1) = -2\sin(2x - 1)$ and $\frac{d}{dx}(4 - 5x)^6 = -30(4 - 5x)^5$.

Powers of trigonometric functions and longer chains

The notation $\sin^2 x$ means $(\sin x)^2$: the inner function is $\sin x$ and the outer function is squaring. This is different from $\sin(x^2)$, where the inner function is $x^2$ and the outer function is sine. The two are differentiated differently, and confusing them is a common source of error.

A chain can also have more than two links. For $y = e^{\sin(2x)}$, the input is doubled, then the sine is taken, then the exponential. The chain rule applies once for each link, working from the outside in: differentiate the outermost function, then multiply by the derivative of the next function in, and so on.

Tangents and gradients with the chain rule

Once a derivative is found, it is used exactly as in 5.4: substitute a value of $x$ to find the gradient, then use the point–gradient form of a line.

Finding unknown constants

When a function contains unknown constants, each given value of the function or of its gradient provides one equation. Two unknowns need two pieces of information.

The product rule

A product is not differentiated factor by factor

The derivative of a sum is the sum of the derivatives, but the derivative of a product is not the product of the derivatives. A single example shows this: $x \cdot x = x^2$ has derivative $2x$, but the product of the derivatives of $x$ and $x$ is $1 \times 1 = 1$.

To see the correct rule, think of $y = uv$ as the area of a rectangle with sides $u$ and $v$, where both sides depend on $x$. When $x$ increases by $\delta x$, the sides increase by $\delta u$ and $\delta v$.

The increase in area is

Dividing by $\delta x$:

As $\delta x \to 0$, the last term tends to $0 \times \frac{dv}{dx} = 0$, and the other ratios tend to derivatives. The result is the product rule.

This form is printed in the formula booklet. The order of the two terms does not matter, since they are added.

Using the product rule

Set out a product rule calculation by writing down $u$, $v$ and their derivatives before substituting. This makes the work easy to check and avoids losing a term.

Factorising the result

A derivative is usually wanted for a purpose: to find where the gradient is zero, or to decide where a function increases. Both need the derivative in factorised form. Products involving $e^{kx}$ are especially convenient, because $e^{kx}$ appears in both terms and is never zero.

The product rule often combines with the chain rule, because one or both factors may be composite.

The quotient rule

The rule and where it comes from

A quotient $\frac{u}{v}$ is the product $u \times v^{-1}$, so it could always be differentiated with the product and chain rules. Doing this once, in general, gives a formula that saves time. With $y = uv^{-1}$, the chain rule gives $\frac{d}{dx}\left(v^{-1}\right) = -v^{-2}\frac{dv}{dx}$, so

The quotient rule is in the formula booklet in this form. Unlike the product rule, the order of the terms in the numerator matters, because they are subtracted: the term with $\frac{du}{dx}$, the derivative of the numerator, comes first.

Using the quotient rule

The factorised answers say more than the unsimplified ones. In (a), the derivative $-\frac{7}{(x - 3)^2}$ is negative for every $x \neq 3$, so the graph of $y = \frac{2x + 1}{x - 3}$ decreases on each side of its vertical asymptote. In (c), every factor of the derivative is positive or zero, so $y = \frac{e^x}{x^2 + 1}$ never decreases; its gradient is zero only at $x = 1$.

The derivative of tan x

The quotient rule gives the derivative of $\tan x = \frac{\sin x}{\cos x}$, a standard result that HL students use as a formula in 5.15.

Quotient rule or not?

The quotient rule always works for a quotient, but it is not always the quickest method.

For example, the quotient rule applied to $\frac{4}{(x^2 + 1)^2}$ gives the correct answer, $-\frac{16x}{(x^2 + 1)^3}$, but only after a longer calculation in which the term $(x^2 + 1)^2 \times 0$ has to be handled. The chain rule reaches it in one line.

Combining the rules

Choosing the rule from the structure

Any function built from the standard functions can be differentiated by identifying its overall structure, applying the matching rule, and then dealing with each piece in the same way. The overall structure is the last operation that would be carried out when evaluating the function.

Inside each rule, any piece that is itself composite or a product needs its own rule. A product of composites, such as $e^{2x}\sin(3x)$, needs the product rule on the outside and the chain rule for each factor.

Using technology with combined rules

Some equations $f'(x) = 0$ cannot be rearranged neatly, and a GDC finds their solutions from the graph of $y = f'(x)$. A graphic display calculator (GDC) can also evaluate the gradient at a point numerically, which gives a quick check on a derivative found by hand: the two values should agree.

Simplifying with logarithm laws first

When a logarithm contains a product, a quotient or a power, the laws of logarithms can split it into simpler terms before differentiating. This is an example of equivalence: the two forms are the same function, but one is much easier to differentiate.

Rates of change in context

Interpreting a derivative as a rate

If a quantity $Q$ depends on a variable $t$, then $\frac{dQ}{dt}$ is the rate at which $Q$ changes as $t$ increases. Its units are the units of $Q$ divided by the units of $t$, and its sign shows whether $Q$ is increasing or decreasing (see 5.1). The standard derivatives and the chain, product and quotient rules make it possible to find such rates for the exponential, trigonometric and rational models used in science and economics.

Air pressure decreases with height above sea level. A simple model, close to standard values in the lower atmosphere, is

where $p$ is the pressure in kilopascals (kPa) and $h$ is the height in kilometres. By the chain rule,

At sea level the pressure falls by about $12.2$ kPa for each kilometre climbed; at a height of $5$ km, $p'(5) = -12.156e^{-0.6} \approx -6.67$ kPa per km. The rate of decrease is proportional to the pressure itself, which is the property of exponential models: $p'(h) = -0.12\,p(h)$.

Linking rates with the chain rule

The chain rule also connects rates of change of different quantities. If $A$ depends on $r$, and $r$ depends on time $t$, then

So the rate of change of $A$ can be found from the rate of change of $r$, even without writing $A$ directly in terms of $t$. HL students develop this idea further as related rates of change in 5.14.

Product and quotient models

Many models multiply a growing factor by a decaying one. Sales of a new product, for example, often rise quickly after its launch and then fall away as the market becomes saturated.

Quotient models often describe a quantity that approaches a limit. A simple model of learning, for example, gives the number of words $N$ of a new vocabulary list that a student can recall after $t$ minutes of study as

The quotient rule, with $u = 40t$ and $v = t + 5$, gives

At the start, $N'(0) = 8$ words per minute; after 5 minutes, $N'(5) = 2$; after 15 minutes, $N'(15) = 0.5$. The rate is always positive but falls steadily, and $N$ approaches $40$, the length of the list. This pattern of diminishing returns is typical of quotient models with a horizontal asymptote (see 2.8).

Exam-style questions on the chain, product and quotient rules

Typical longer questions ask for a derivative, then use it: to find a gradient, a tangent or normal, or the points where the gradient is zero. A context question adds units and interpretation.

Self-check

• To do:

  I can differentiate rational powers of $x$, including roots and reciprocals of roots.

• To do:

  I can state and use the derivatives of $\sin x$, $\cos x$, $e^x$ and $\ln x$, and I know why radians are needed.

• To do:

  I can identify the inner and outer functions of a composite function and apply the chain rule, including chains of three functions.

• To do:

  I can tell $\sin^2 x$ from $\sin(x^2)$ and differentiate each.

• To do:

  I can use the product rule and factorise the result to find where the derivative is zero.

• To do:

  I can use the quotient rule with the numerator terms in the right order, and decide when a simpler method is better.

• To do:

  I can combine the rules, simplifying with the laws of logarithms where possible.

• To do:

  I can find and interpret rates of change of real quantities, with units.

(b) Write $y = 6x^{\frac{1}{2}}$. Then

(c) Write $y = 4x^{-\frac{1}{3}}$. Subtracting 1 from $-\frac{1}{3}$ gives $-\frac{4}{3}$:

(d) Divide each term of the numerator by $\sqrt{x} = x^{\frac{1}{2}}$ first:

(b) Use $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2} = 1$:

This is about $4.26$.

(b) The inner function $-2x$ has derivative $-2$: $\frac{dy}{dx} = -2e^{-2x}$.

(c) With $u = x^2 + 1$, $\frac{dy}{du} = \frac{1}{u}$ and $\frac{du}{dx} = 2x$:

(d) Write $y = (1 - 4x)^{\frac{1}{2}}$. The outer function is a power, and the inner function has derivative $-4$:

(e) Write $y = (2x + 3)^{-3}$:

(Using the double angle identity from 3.6, this equals $\sin 2x$.)

(b) The outer function is sine and the inner function is $x^2$:

(c) Three links: $e^{(\dots)}$, then $\sin(\dots)$, then $2x$.

(d) $y = \left(\sin(4x)\right)^3$. Three links: cubing, then sine, then $4x$.

Write $y = (2x + 5)^{\frac{1}{2}}$ and use the chain rule:

At $x = 2$ the gradient is $\frac{1}{3}$.

Tangent: $y - 3 = \frac{1}{3}(x - 2)$, so $y = \frac{1}{3}x + \frac{7}{3}$, or $x - 3y + 7 = 0$.

Normal: gradient $-3$, so $y - 3 = -3(x - 2)$, giving $y = -3x + 9$.

(b) $f'(x) = -6e^{-2x}$. Solve $-6e^{-2x} = -1$:

This is about $0.896$.

(b) Let $u = x$ and $v = \ln x$. Then $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \frac{1}{x}$.

(c) Let $u = x^3$ and $v = e^{2x}$. Then $\frac{du}{dx} = 3x^2$ and, by the chain rule, $\frac{dv}{dx} = 2e^{2x}$.

Both terms share the factor $x^2e^{2x}$, so

which is the given result.

(b) A horizontal tangent has gradient $0$. Since $e^{-x} > 0$ for all $x$,

$f(0) = 0$ and $f(2) = 4e^{-2}$. The points are $(0, 0)$ and $\left(2, \frac{4}{e^2}\right)$.

(b) $u = \sin x$, $v = x$, so $\frac{du}{dx} = \cos x$ and $\frac{dv}{dx} = 1$.

(c) $u = e^x$, $v = x^2 + 1$, so $\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = 2x$.