5.9 Further differentiation

Learning objectives

• Find the derivative of $x^{n}$ for any rational power $n$, after rewriting roots and fractions as powers.

• Write down the derivatives of $\sin x$, $\cos x$, $\tan x$, $\mathrm{e}^{x}$ and $\ln x$, with $x$ in radians for the trigonometric functions.

• Find the derivative of a composite function using the chain rule, including $\mathrm{e}^{kx}$, $\sin(kx)$, $\cos(kx)$ and $\ln(kx)$

• Find the derivative of a product of two functions using the product rule.

• Find the derivative of a quotient of two functions using the quotient rule, or the product rule as an alternative.

• Determine which rule, or combination of rules, a function needs, and check a derivative with the GDC.

• Use these derivatives to find gradients, tangents, stationary points and optimal values in context.

• Solve related rates of change problems by linking rates with the chain rule, and interpret the result with its units.

Introduction

The power rule of subtopic 5.3, Differentiating polynomials, handles functions built from whole-number powers of $x$. Most models in science, economics and medicine are built from other pieces: square roots and fractional powers, exponential growth and decay, logarithms, and the sine and cosine curves of periodic motion. These pieces are also joined together in ways that the power rule cannot handle: one function is fed into another, two functions are multiplied, or one is divided by another. Three rules, the chain rule, the product rule and the quotient rule, together with five standard derivatives, are enough to differentiate every such function.

The chain rule has a second use. When two quantities are linked by a formula and one of them changes with time, the chain rule gives how fast the other changes. The wind chill index used by weather services in Canada, the United States and the United Kingdom is a formula of this kind: when the wind strengthens, the chain rule gives how quickly the temperature "felt" by the skin is falling.

Derivatives of rational powers

Extending the power rule

Subtopic 5.3 established the power rule for integer powers: if $f(x) = ax^{n}$, then $f'(x) = anx^{n-1}$

A numerical check makes the rule believable for $n = \frac{1}{2}$. The gradient of $y = \sqrt{x}$

The power rule predicts $f'(x) = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$

Rewriting before differentiating

The power rule only works on a term of the form $ax^{n}$. Roots, fractions with $x$ in the denominator, and products of powers must first be rewritten with the laws of exponents from subtopics 1.5 and 1.10.

The exponent of the derivative is always one less than the original exponent, and subtracting 1 from a fraction is where most slips happen: $\frac{1}{2} - 1 = -\frac{1}{2}$ and $-\frac{1}{3} - 1 = -\frac{4}{3}$

Domains of rational powers

A power such as $x^{\frac{1}{2}}$ is defined only for $x \ge 0$, and its derivative $\frac{1}{2\sqrt{x}}$

Standard derivatives

Five functions besides $x^{n}$ are differentiated so often that their derivatives are learned as facts. All the other functions met here are differentiated by combining them with the chain, product and quotient rules.

The exponential function

Every exponential function $y = a^{x}$ with $a > 1$ grows at a rate proportional to its own value. The gradient of $y = a^{x}$

There is exactly one base, between 2.5 and 3, for which the gradient at $x = 0$ is exactly 1. That base is Euler's number $\mathrm{e} = 2.718\,28\ldots$, met in subtopic 1.5. The gradient of any exponential function at a point $x$ is its value there multiplied by its gradient at 0, because a chord from $x$ to $x + h$

The second factor is the gradient of the chord from 0 to $h$, which approaches the gradient at 0 as $h$ shrinks. For $a = \mathrm{e}$ that gradient is 1, so the function $\mathrm{e}^{x}$ is its own derivative.

At every point on the curve $y = \mathrm{e}^{x}$, the gradient equals the height. At $x = 0$ both are 1; at $x = 3$ both are $\mathrm{e}^{3} \approx 20.1$

The natural logarithm

The natural logarithm $\ln x$ is the inverse of $\mathrm{e}^{x}$, so the graph of $y = \ln x$ is the reflection of $y = \mathrm{e}^{x}$

The graph of the two functions shows the reflection: the tangent to $y = \mathrm{e}^{x}$ at $(1, \mathrm{e})$ has gradient $\mathrm{e}$, and its mirror image, the tangent to $y = \ln x$

The derivative $\frac{1}{x}$ is positive for every $x > 0$, so $\ln x$ is always increasing. The gradient $\frac{1}{x}$

Sine and cosine

The gradient of $y = \sin x$ can be read from its graph. At $x = 0$ the curve climbs most steeply; at $x = \frac{\pi}{2}$

The size of the gradient at 0 depends on the angle unit. With $x$ in radians, $\frac{\sin x}{x}$ gets closer and closer to 1 as $x$ approaches 0, so the curve leaves the origin with gradient exactly 1.

That gradient of 1 at the origin makes the derivative of $\sin x$ exactly $\cos x$, with no extra constant. The same reasoning applied to $\cos x$, whose graph is the sine curve moved $\frac{\pi}{2}$

Tangent

The derivative of $\tan x = \frac{\sin x}{\cos x}$ follows from the quotient rule, applied to this fraction. The result, $\frac{1}{\cos^{2} x}$

The chain rule

Composite functions: inner and outer functions

A function such as $y = (3x^{2} - 5)^{4}$ cannot be differentiated term by term without first expanding a fourth power, and $y = \sin(2x + 5)$

The second function in the chain acts on the output of the first. Naming the inner function $u$ turns the composite function into two simple functions, each of which has a standard derivative.

A reliable way to find the outer function is to ask which operation would be done last when the function is evaluated on a calculator. For $\sqrt{25 - x^{2}}$ the square root is pressed last, so the square root is the outer function.

Why the rates multiply

A composite function is a chain of two machines: $x$ goes into the inner machine and comes out as $u$, and $u$ goes into the outer machine and comes out as $y$. Each machine has its own rate of change. If $u$ changes 3 times as fast as $x$, and $y$

The same idea in symbols: a small change $\Delta x$ produces a change $\Delta u$ in the inner function, which produces a change $\Delta y$ in the outer function. Writing the overall ratio as a product of two ratios and letting $\Delta x$ shrink to 0 gives the rule.

The chain rule method

1. Identify the inner function and call it $u$; write $y$ as a function of $u$.

2. Find $\frac{\mathrm{d}y}{\mathrm{d}u}$

With practice, steps 1 to 4 merge into one line: "derivative of the outside, with the inside left alone, times derivative of the inside". Writing out $u$ is still the safest layout in a long question, and it shows the method clearly.

The 3Blue1Brown video shows the chain rule as a small nudge passing through two functions in turn, as in the function-machine diagram, and the product rule as the growth of a rectangle's area.

YouTube

Linear inner functions: e^{kx}, sin(kx), cos(kx) and ln(kx)

When the inner function is $u = kx$ (or $kx + b$), its derivative is the constant $k$, and the chain rule simply multiplies the standard derivative by $k$. These results are used constantly in models, so they are worth knowing at sight.

The result for $\ln(kx)$ can look wrong at first, but the laws of logarithms from subtopic 1.9 explain it: $\ln(kx) = \ln k + \ln x$, and $\ln k$ is a constant whose derivative is 0. Multiplying $x$

General forms: e^{f(x)} and ln f(x)

Two further patterns come from the chain rule with any inner function $f(x)$.

For example, $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{x^{2}}\right) = 2x\mathrm{e}^{x^{2}}$

Applying the chain rule more than once

Some functions are chains of three or more functions. The chain rule extends by multiplying one rate for each link. For $y = \sin^{2}(3x) = \left(\sin(3x)\right)^{2}$, the last operation is squaring, inside that is the sine, and inside that is $3x$

Each factor is the derivative of one layer, working from the outside in, with everything inside that layer left unchanged.

The product rule

Differentiating a product

The derivative of a product is not the product of the derivatives. For $y = x^{2} \times x^{3} = x^{5}$, the derivative is $5x^{4}$

A picture of area explains the rule. A rectangle with sides $u$ and $v$ has area $y = uv$. If both sides grow, $u$ by a small amount $\Delta u$ and $v$

Dividing by $\Delta x$ and letting $\Delta x$ shrink to 0 makes the corner term vanish, because it is a product of two small quantities divided by one small quantity, and so tends to 0. What remains is the product rule.

The check with $x^{2} \times x^{3}$ confirms it: $u = x^{2}$, $v = x^{3}$

Setting out the product rule

A clear layout lists the four ingredients before combining them, so each derivative is found separately and nothing is lost.

Then the rule $u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}$

Products with an exponential factor

Models of a quantity that rises and then dies away often have the form $y = at\,\mathrm{e}^{-bt}$: the factor $at$ makes the quantity grow at first, and the decaying exponential eventually pulls it back to 0. The concentration of a drug in the bloodstream after a single dose is a standard example. Taking out the exponential factor in the derivative shows at once where the maximum is.

The quotient rule

Deriving the quotient rule

A quotient $y = \frac{u}{v}$ is the product $u \times v^{-1}$. The product rule, with the chain rule for

The subtraction makes the order important, unlike in the product rule. Swapping the two terms changes the sign of the whole answer. A memory aid is "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared". The denominator $v^{2}$ is best left unexpanded, because leaving it squared keeps the answer factorised and shows at once where the derivative is undefined.

The derivative of tan x

The quotient rule proves the standard derivative of $\tan x$. Let $u = \sin x$ and $v = \cos x$, so $\frac{\mathrm{d}u}{\mathrm{d}x} = \cos x$

The last step uses $\sin^{2} x + \cos^{2} x = 1$, true for every angle because $(\cos x, \sin x)$

The product rule as an alternative

Any quotient can be written as a product with a negative power, so the product rule (with the chain rule) can always replace the quotient rule. For $y = \frac{\ln x}{x} = \ln x \times x^{-1}$

The quotient rule gives the same result in one step: $\frac{x \times \frac{1}{x} - \ln x \times 1}{x^{2}} = \frac{1 - \ln x}{x^{2}}$

Optimisation with the quotient rule

The optimisation method of subtopic 5.7 stays the same: build the objective function in one variable from a constraint, differentiate, solve $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, justify the nature of the point and interpret. What changes is that the objective function may now be a quotient, a root or an exponential, and the quotient rule often leads to a numerator that can be set to 0 and solved exactly.

The graph of the area function confirms the result: $A$ rises steeply for small angles, peaks at $\theta = 1$, and falls slowly as the bed becomes a nearly full circle of small radius.

Combining the rules

Choosing the rule

Many functions need two or three rules. The rule to apply first is decided by the last operation in the function, the one that would be carried out last if it were evaluated. That operation's rule gives the overall structure, and the other rules are used inside it to find the pieces.

Simplifying before differentiating saves work where it is possible. The laws of logarithms turn $\ln(x\mathrm{e}^{x})$ into $\ln x + x$, with derivative $\frac{1}{x} + 1$

Checking a derivative with the GDC

A GDC without a computer algebra system cannot differentiate symbolically, but it can find the numerical value of a derivative at a point, which is the quickest check of an algebraic derivative. The method is to evaluate the algebraic answer at some convenient value of $x$ and compare it with the GDC's numerical derivative of the original function at the same value.

• On a TI-84 Plus CE, the numerical derivative is MATH → 8:nDeriv(; on a graph, 2nd TRACE (CALC) → 6:dy/dx gives the gradient at a traced point.

• On a Casio fx-CG50, in the Run-Matrix menu, OPTN → CALC → d/dx gives the numerical derivative.

• With trigonometric functions the GDC must be in radian mode.

For $y = \mathrm{e}^{0.5x}\sin x$, the product rule gives $\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{0.5x}(0.5\sin x + \cos x)$

Related rates of change

Rates linked by a formula

In many situations two or more quantities change at the same time and are linked by a formula. When a spherical balloon is inflated, its volume $V$ and radius $r$ both increase with time $t$, and $V = \frac{4}{3}\pi r^{3}$

The connection is the chain rule with time as the original variable. Time $t$ is the input, $x$ depends on $t$, and $y$ depends on $x$: the inner function is $x(t)$ and the outer function is $y(x)$

The formula also works in reverse. If $\frac{\mathrm{d}y}{\mathrm{d}t}$ is known and $\frac{\mathrm{d}x}{\mathrm{d}t}$

The method for related rates

1. Draw a diagram if the situation is geometric, and name every quantity that changes with a letter. Keep fixed quantities as numbers.

2. Write down the rate that is given, with its sign and units (a quantity that is decreasing has a negative rate), and the rate that is wanted.

3. Find an equation linking the two quantities whose rates are involved. If a third variable appears, remove it with another relation (similar triangles, Pythagoras, a given constraint).

4. Differentiate the equation with respect to the variable it contains, to get $\frac{\mathrm{d}y}{\mathrm{d}x}$.

5. Combine with the chain rule, $\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}$

Eliminating a variable

A cone filling with water has three changing quantities: the volume $V$, the depth $h$ and the radius $r$ of the surface. The formula $V = \frac{1}{3}\pi r^{2}h$

The Khan Academy video works through a problem of this type, a cone of water being filled, and shows how the similar triangles remove the radius before differentiating.

YouTube

Related rates in other contexts

The same method applies whenever a formula links two changing quantities, not only in geometry. In electricity, the voltage of an alternating current supply varies sinusoidally with time, and the rate of change of the voltage is found with the chain rule on $\sin(\omega t)$, where $\omega$ is the angular frequency. Mains electricity in the United Kingdom has frequency 50 Hz and a nominal voltage of 230 V, with peak voltage $230\sqrt{2} \approx 325$

A rate along a curve

A related rate can also come from a curve given as a graph rather than a geometric formula. If a point moves along the curve $y = f(x)$, its vertical and horizontal rates of change are linked by the gradient of the curve at that point: $\frac{\mathrm{d}y}{\mathrm{d}t} = f'(x) \times \frac{\mathrm{d}x}{\mathrm{d}t}$

Case study: The wind chill index

Wind makes cold air feel colder, because moving air carries heat away from exposed skin faster than still air. In November 2001, Canada, the United States and the United Kingdom adopted a new wind chill index, based on a model of heat loss from a bare face walking into the wind. The index is approximated, to within one degree, by a formula in the air temperature $T_a$ (°C) and the wind speed $v$ (km/h) measured at the standard height of 10 m.

Source: Wikipedia, "Wind chill" (the 2001 Joint Action Group for Temperature Indices formula), accessed 2026.

The index is defined only for air temperatures of 10 °C or below and wind speeds above 4.8 km/h. The power $0.16$ is a rational exponent, so the formula is differentiated with the extended power rule. As a check of the formula, at $T_a = -20$ °C it gives $-24.3$ at 5 km/h and $-32.6$ at 30 km/h, which round to the published values of

Wind chill at a fixed air temperature

On a day when the air temperature is $-15$ °C, the formula becomes a function of wind speed alone. Collecting the constant terms and the $v^{0.16}$ terms:

The derivative gives the change in the felt temperature per extra km/h of wind.

The derivative is negative at every wind speed, so stronger wind always makes it feel colder. Its size shrinks quickly, because the power $-0.84$ is negative. Going from calm to a light wind makes a large difference, while an extra km/h in a gale changes the felt temperature by less than a tenth of a degree.

A strengthening wind

Suppose the wind speed is 20 km/h and increasing at 6 km/h every hour as a storm approaches, with the air temperature steady at $-15$ °C. The wind chill changes with time because the wind speed does, so the chain rule links the two rates.

At that moment the felt temperature is falling at about 1.34 °C per hour. The same increase in wind at 40 km/h would lower the wind chill by only $0.125 \times 6 \approx 0.750$ °C per hour, so for the same rise in wind speed per hour, the felt temperature drops faster while the wind is still light.

Limitations of the model

The formula is an approximation, accurate to about one degree, of a more detailed heat-loss model. It assumes a bare face in an open field, walking into the wind at about 5 km/h, and corrects the wind speed measured at 10 m to the speed at face height. It holds only for $T_a \le 10$ °C and $v > 4.8$ km/h, so its derivative should not be used at very low wind speeds, where $v^{-0.84}$

Exam focus

Further differentiation is examined on HL Papers 1 and 2, and it supports many of the calculus questions on Paper 3. Short Paper 1 questions ask you to find a derivative (2–3 marks), often followed by a gradient, a tangent or a stationary point. Paper 2 and Paper 3 questions use the rules inside longer models: show that a derivative has a given form (so that a later part can use it), hence find a maximum, or show that a rate is always positive (which needs a reason, R1). Related rates questions are usually 4–6 marks and are not always labelled as such: a phrase like "the radius is increasing at 3 mm per second" is the signal.

Markschemes give (M1) for an attempt at the right rule (the correct structure of the chain, product or quotient rule, even with a slip), and A1 for each correct part. In related rates, the given rate is often worth (A1) on its own, the chain rule product (M1), and the final value with units and sign A1. When a derivative leads to an equation that cannot be solved by hand, solve it on the GDC and write down the equation you solved.

The commonest losses in further differentiation come from the chain rule left incomplete inside another rule: $\frac{\mathrm{d}}{\mathrm{d}t}\mathrm{e}^{-0.2t}$ written as $\mathrm{e}^{-0.2t}$

Related rates questions reward a clear statement of each rate before any calculation. Writing "$\frac{\mathrm{d}v}{\mathrm{d}t} = 6$" and "$\frac{\mathrm{d}W}{\mathrm{d}t} = \frac{\mathrm{d}W}{\mathrm{d}v} \times \frac{\mathrm{d}v}{\mathrm{d}t}$

Practice questions

1. Let $f(x) = 5x^{\frac{2}{5}}$, $x > 0$

Summary

• For any rational $n$, $\frac{\mathrm{d}}{\mathrm{d}x}(x^{n}) = nx^{n-1}$

Self-check

• To do:

  I can rewrite $\sqrt{x}$, $\frac{1}{\sqrt{x}}$