5.3 Differentiating polynomials

Introduction

Subtopic 5.1, Limits and the derivative, describes the derivative as the gradient of a curve at each point, found as the limit of the gradients of chords. That idea is clear but slow: estimating one gradient needs a table of chord gradients. For the functions used most often in modelling, sums of powers of $x$, a single rule gives the whole gradient function at once. With it, a cost function becomes a marginal cost, a distance becomes a speed and a volume becomes a rate of growth, in one line of algebra.

The rule comes from a pattern in the gradients of $x^2$, $x^3$, $x^4$, …, and the same pattern turns out to hold for negative powers such as $x^{-1} = \frac{1}{x}$. Together with two simple facts, that multiplying a function by a number multiplies its gradient by that number and that the gradient of a sum is the sum of the gradients, it allows any function built from integer powers to be differentiated term by term. Every later subtopic of calculus at this level, from tangents to optimisation, starts from this skill.

The power rule

A pattern in the gradients of powers of x

The gradient of $y = x^2$ at $x = 3$ can be estimated from chords, as in subtopic 5.1. A chord from $(3, 9)$ to a nearby point $(3 + h, (3 + h)^2)$ has gradient $\frac{(3 + h)^2 - 9}{h}$. As $h$ shrinks, these chord gradients close in on the gradient of the tangent. The same calculation for $y = x^3$ at $x = 2$ gives a second column.

Repeating the calculation at other points and for other powers fills in a table of gradients. Each row shows the gradient of the curve at $x = 1$, $2$ and $3$, and the last column the formula that fits every value in the row.

For $x^3$ the gradients 3, 12, 27 are three times the squares 1, 4, 9, so the gradient function is $3x^2$. For $x^4$ the gradients 4, 32, 108 are four times the cubes 1, 8, 27, giving $4x^3$. In each row the old power comes down in front as a multiplier and the new power is one less than the old one.

In Leibniz notation the same rule reads $\frac{\mathrm{d}}{\mathrm{d}x}\left(ax^n\right) = anx^{n-1}$. The symbol $\frac{\mathrm{d}}{\mathrm{d}x}$ in front of an expression is an instruction: differentiate what follows with respect to $x$. So $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^4\right) = 4x^3$, and if $y = x^4$ then $\frac{\mathrm{d}y}{\mathrm{d}x} = 4x^3$; both say the same thing.

Why the power comes down

The tangents to $y = x^2$ drawn at $x = -1$, $1$ and $2$ have gradients $-2$, $2$ and $4$, exactly $2x$ at each point. The tangent at a negative $x$ slopes down and the gradient formula gives a negative value there, so one formula describes the steepness on both sides of the vertex.

The pattern can be justified for every positive integer power with a short piece of algebra. The chord gradient of $y = x^n$ between $x$ and $x + h$ is $\frac{(x + h)^n - x^n}{h}$, and the key step is expanding $(x + h)^n$.

The 3Blue1Brown video "Derivative formulas through geometry" makes the same argument with pictures: a square of side $x$ grows by two thin strips of area $x \times \mathrm{d}x$ when its side grows a little, and a cube grows by three thin slabs, which is where the 2 in $2x$ and the 3 in $3x^2$ come from.

The powers 1 and 0

Two special cases of the power rule describe straight lines. The line $y = x$ is $x^1$, and the power rule gives $1 \times x^0 = 1$: the line has gradient 1 everywhere, as its graph confirms. More generally $\frac{\mathrm{d}}{\mathrm{d}x}(mx) = m$, the familiar gradient of a straight line from subtopic 2.1.

A constant such as $y = 7$ can be written $7x^0$, because $x^0 = 1$ for $x \ne 0$. The power rule then gives $7 \times 0 \times x^{-1} = 0$. The graph of $y = 7$ is a horizontal line, which has gradient 0 at every point, so the rule and the picture agree.

A constant added to a function moves its graph up or down without changing its shape, so it changes no gradients. The curves $y = x^2$, $y = x^2 + 5$ and $y = x^2 - 3$ are vertical translations of one another and all have gradient function $2x$. The constant term of a polynomial therefore disappears when it is differentiated.

Constant multiples and sums

The constant multiple rule

Multiplying a function by 3 stretches its graph vertically by a factor of 3. Every point moves to three times its height, so every rise between two points becomes three times as large while every run stays the same. Every chord gradient is multiplied by 3, and so is every tangent gradient. In symbols, the chord gradient of $3f$ is $\frac{3f(x + h) - 3f(x)}{h} = 3 \times \frac{f(x + h) - f(x)}{h}$, and the factor 3 is still there in the limit.

The coefficient $a$ in the power rule $ax^n \Rightarrow anx^{n-1}$ is exactly this rule at work: differentiate $x^n$ to get $nx^{n-1}$, then multiply by $a$. The rule holds for any constant, including fractions and negative numbers, and it explains why $\frac{x^4}{8}$ has derivative $\frac{4x^3}{8} = \frac{x^3}{2}$.

The sum rule

When two functions are added, their heights add at every $x$, and so do their rises over any run. The chord gradient of $f + g$ splits into two pieces:

As $h \to 0$, the first piece tends to $f'(x)$ and the second to $g'(x)$. Subtraction works the same way, because subtracting $g$ is adding $-1 \times g$.

Differentiating a polynomial term by term

The three rules together differentiate any polynomial: each term $ax^n$ becomes $anx^{n-1}$, the constant term becomes 0, and the results are added. Because every power drops by 1, the derivative of a polynomial of degree $n$ is a polynomial of degree $n - 1$: a cubic has a quadratic derivative, a quadratic a linear one, and a linear function a constant one.

Writing the derivative one term at a time, in the same order as the original, keeps the work easy to check. Each coefficient should be simplified to a single number.

Fractional coefficients

Models often have fractional coefficients, such as $\frac{x^4}{8}$ or $-\frac{2}{3}x^3$. The power rule treats them like any other number: multiply the coefficient by the power and simplify. A fraction written over a number, such as $\frac{x^4}{8}$, is the same as $\frac{1}{8}x^4$, so its derivative is $\frac{1}{8} \times 4x^3 = \frac{x^3}{2}$.

Negative integer powers

Rewriting fractions as powers

Functions such as $\frac{3}{x}$, $\frac{128}{x}$ and $\frac{5}{x^2}$ appear whenever one quantity is divided by another: the height of a box of fixed volume, the average cost per item, the time for a journey at a given speed. None of these is written as a power of $x$, so the power rule cannot be applied to them as they stand. The negative exponent rule $x^{-k} = \frac{1}{x^k}$ from subtopic 1.5 rewrites each one as a power.

Only the part of the term that is in the denominator moves into the power; a number in the denominator stays in the coefficient as a fraction. The table shows the commonest rewritings.

Why the power rule still works

The power rule was found from positive powers, so it needs checking for $n = -1$. The chord gradient of $y = \frac{1}{x}$ between $x$ and $x + h$ can be simplified by putting the two fractions over a common denominator.

As $h \to 0$ this tends to $-\frac{1}{x^2} = -x^{-2}$, which is exactly what the power rule gives for $x^{-1}$: bring down the $-1$ and reduce the power to $-2$. Numerically, the chord gradients at $x = 2$ for $h = 1, 0.1, 0.01, 0.001$ are $-0.1667$, $-0.2381$, $-0.2488$ and $-0.2499$, closing in on $-\frac{1}{4}$. The same kind of argument works for every negative integer power.

The graph of $y = \frac{1}{x}$ falls from left to right on each side of the vertical asymptote $x = 0$, so every tangent has a negative gradient, and $-\frac{1}{x^2}$ is negative for every $x \ne 0$. The tangents are steepest close to the asymptote, where $x^2$ is small and $\frac{1}{x^2}$ large, and almost flat far from it.

The function $\frac{1}{x}$ is not defined at $x = 0$, and neither is its derivative. A function containing a negative power always comes with the condition $x \ne 0$, and its derivative carries the same condition.

Subtracting 1 from a negative power

Reducing a negative power by 1 makes it more negative: $-1$ becomes $-2$, $-2$ becomes $-3$, and $-5$ becomes $-6$. The coefficient also picks up a negative factor, so a positive term $\frac{a}{x^k}$ always has a negative derivative $-\frac{ka}{x^{k+1}}$, and a negative term has a positive one.

A negative power in physics: gravity and height

Gravity weakens with distance from the Earth's centre according to an inverse-square law. At a distance $r$ metres from the centre, outside the Earth, the gravitational field strength is modelled by $g(r) = \frac{GM}{r^2}$ in $\text{m s}^{-2}$, where $GM \approx 3.986 \times 10^{14}\text{ m}^3\text{ s}^{-2}$ is a measured constant for the Earth. Rewritten as a power, $g(r) = GM\,r^{-2}$, the model can be differentiated with respect to $r$ by the power rule.

The derivative is negative for every $r$, so the field strength falls as the distance grows, and the size of the derivative is the loss of field strength per metre of extra height. Because $r^3$ grows quickly, the rate of loss becomes smaller further from the Earth.

Sources: $GM$ for the Earth (geocentric gravitational constant) and mean radius 6,371 km, Wikipedia, "Standard gravitational parameter" and "Earth radius", accessed 2026.

Rewriting products and quotients before differentiating

Products: expand first

The rules so far apply to sums of powers. A product such as $(2x - 1)(x^2 + 3)$ is not a sum, and there is no rule at this level for differentiating a product directly. Expanding the brackets turns it into a polynomial, which the power rule handles term by term.

It is tempting to differentiate each factor and multiply the results, but this gives the wrong answer. The simplest case shows why: $x^2 \times x^3 = x^5$, whose derivative is $5x^4$, while differentiating the factors separately gives $2x \times 3x^2 = 6x^3$. The two expressions are not equal, so the derivative of a product is not the product of the derivatives.

Quotients with a single-term denominator: split first

A quotient whose denominator is a single term, such as $\frac{x^3 - 4x + 8}{2x}$, can be split into separate fractions, one for each term of the numerator. Each fraction then simplifies by the laws of exponents to a single power of $x$.

A denominator with more than one term, such as $\frac{1}{x + 1}$, cannot be split in this way: $\frac{1}{x + 1}$ is not $\frac{1}{x} + 1$. Functions of that kind need the quotient or chain rule and are not differentiated at this level.

Derivatives with other variables and parameters

Other letters for the variables

Models use letters that suggest their meaning: $V$ for volume, $r$ for radius, $s$ for distance, $t$ for time, $C$ for cost, $q$ for quantity. The rules of differentiation do not depend on the letters. If $V = \frac{4}{3}\pi r^3$, the derivative of $V$ with respect to $r$ is written $\frac{\mathrm{d}V}{\mathrm{d}r}$ and found by the power rule with $r$ in place of $x$.

The notation $\frac{\mathrm{d}V}{\mathrm{d}r}$ is read "the derivative of $V$ with respect to $r$", and it is the rate of change of volume as the radius grows, measured in units of volume per unit of length, for example $\text{cm}^3$ per cm. The result $4\pi r^2$ is also the surface area of the sphere: a thin layer added to the outside of a sphere has a volume of about its surface area times its thickness.

Free fall: from distance to speed

An object dropped from rest near the Earth's surface, with air resistance ignored, falls a distance $s = \frac{1}{2}gt^2$ metres in the first $t$ seconds, where $g$ is the acceleration due to gravity. Taking $g \approx 9.8\text{ m s}^{-2}$ (standard gravity is defined as $9.80665\text{ m s}^{-2}$) gives $s = 4.9t^2$. Differentiating distance with respect to time gives the instantaneous rate of change of distance, the speed.

The speed grows by $9.8\text{ m s}^{-1}$ every second, which is what an acceleration of $9.8\text{ m s}^{-2}$ means. One application of the power rule converts the distance formula into a speed formula.

Parameters are constants

Many models contain a parameter, a letter such as $k$, $a$ or $b$ that stands for a fixed number whose value is not yet known. When differentiating with respect to $x$, a parameter behaves exactly like a number: $kx^3$ has derivative $3kx^2$, $k^2x$ has derivative $k^2$, and a term such as $5k$ or $k^2$ on its own is a constant with derivative 0. The value of the parameter is usually found afterwards from a condition, such as a known gradient at a point.

Two unknown parameters need two conditions. Each condition on the gradient gives one equation, and the pair is solved as simultaneous equations, by hand or with the GDC's simultaneous equation solver (subtopic 1.8).

Gradients at a point and technology

Substituting into the derivative

The derivative $f'(x)$ is a function: it gives the gradient at every value of $x$. The gradient at one particular point $x = a$ is the number $f'(a)$, found by substituting $a$ into the derivative, never into the original function. The value $f(a)$ is the height of the graph at $x = a$, and $f'(a)$ is its steepness there.

A function and its derivative can be drawn one above the other on the same $x$-scale. The cubic $y = x^3 - 3x$ has derivative $y = 3x^2 - 3$. Where the cubic rises, the derivative is above the $x$-axis; where it falls, between $x = -1$ and $x = 1$, the derivative is below it; and at the two turning points the derivative is exactly 0.

Whether $f'(x)$ is positive or negative shows where a function is increasing or decreasing (subtopic 5.2), and where $f'(x) = 0$ the graph is flat (subtopic 5.6, Stationary points). Both start from a derivative found by the rules here.

The GDC's numerical derivative

A GDC does not differentiate algebraically. It estimates the gradient at a single point numerically, by calculating the gradient of a very short chord across the point, much as the tables in subtopic 5.1 do. This gives a fast way to find $f'(a)$ and a reliable check on a derivative found by hand.

1. Enter the function, using the same variable as the calculator (usually $X$).

2. Use the numerical derivative at a value. On a TI-84 Plus CE: MATH → 8:nDeriv(, then enter the expression, the variable and the value. On a Casio fx-CG50: MENU → Run-Matrix, then OPTN → CALC → d/dx. On a TI-Nspire, use the derivative template.

3. Alternatively, graph the function and use the graph's derivative tool (TI-84 Plus CE: 2nd TRACE → 6:dy/dx) at the required $x$.

4. Compare with the hand value. If they differ, check the rewriting of any fractions and the signs of negative powers.

A second check graphs the hand derivative and the calculator's numerical derivative together (on a TI-84 Plus CE, $Y_2 =$ nDeriv($Y_1, X, X$)). If the hand derivative is right, the two graphs lie on top of each other.

Rates in a design: the surface area of an open box

A packaging designer wants an open box (a box with no lid) with a square base of side $x$ cm and a volume of $4000\text{ cm}^3$. The height $h$ is fixed by the volume: $x^2h = 4000$, so $h = \frac{4000}{x^2}$. The box uses card for the base, area $x^2$, and four sides, each of area $xh$, so the area of card in $\text{cm}^2$ is the following function.

The volume condition has produced a negative power, $\frac{16\,000}{x} = 16\,000x^{-1}$, which is how such terms arise in practice. Differentiating gives the rate at which the area of card changes as the base is widened.

At $x = 10$ the derivative is $-140$: widening the base by a small amount reduces the card needed by about $140\text{ cm}^2$ for every centimetre added, because the tall, narrow box gets shorter faster than its base grows. At $x = 30$ the derivative is positive, so widening the base further now increases the card used. At $x = 20$ the derivative is 0, and the table shows the least area of the three, $1200\text{ cm}^2$. Finding such a value systematically, and proving it is the least, is the method of subtopic 5.7, Optimisation; here the derivative already shows in which direction a change in the design helps.

The box is a hypothetical design; all values are calculated from the model.

Exam focus

Differentiation by the power rule is the most frequently tested calculus skill in the course. Almost every calculus question on Paper 1 or Paper 2 opens with find $f'(x)$ or find $\frac{\mathrm{d}y}{\mathrm{d}x}$, usually for 2–3 marks, and then uses the derivative in later parts: a gradient at a point, an equation of a tangent or normal (subtopic 5.4), an interval where the function increases (5.2), or the solution of $f'(x) = 0$ (5.6 and 5.7). The markscheme gives one accuracy mark per correct term of the derivative and withholds the last one if any extra term is present, such as a leftover constant. A term like $\frac{2}{x}$ carries its own method mark for being written as $2x^{-1}$, which can be implied by a correct $-2x^{-2}$ or $-\frac{2}{x^2}$ in the answer.

Other recurring shapes are: a product to expand before differentiating (with a method mark for the expansion), a derivative left in terms of a parameter $k$ and then used to find $k$ from a given gradient, the derivative of a model such as cost, profit, area or volume interpreted as a rate with units, and show that $f'(x) = 0$ leads to a given equation, followed by hence solve it with the GDC.

In part (e) the answer is given, so every step must be written forwards from $A'(x) = 0$; substituting the root back into the cubic proves nothing. In part (f) the GDC does the solving, but the equation it solves must be written down.

Many questions then give a point on the curve, for example "the point $(2, 9)$", and ask for the gradient there or for the equation of the tangent through it (subtopic 5.4). The two coordinates play different roles.

Practice questions

1. Write down $\frac{\mathrm{d}y}{\mathrm{d}x}$ for $y = 4x^6 - \frac{x^3}{3} + 9$. [2]

2. Find $f'(x)$ for $f(x) = \frac{5}{x^2} - \frac{x}{4} + 2$

3. Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ for $y = (3x + 2)^2$

4. The function $g$ is defined by $g(x) = \frac{2x^4 - x + 6}{x^2}$

5. The function $h$ is defined by $h(x) = \frac{x^3}{6} - 4x + \frac{18}{x}$

6. A paper cone has a fixed height of 12 cm. When its base radius is $r$ cm, its volume is $V = 4\pi r^2$ $\text{cm}^3$. (a) Find $\frac{\mathrm{d}V}{\mathrm{d}r}$

7. The curve $y = ax^3 - 4x^2 + 1$ has gradient 8 at the point where $x = 2$. Find the value of $a$

8. Consider the curve $y = x^2 + \frac{16}{x}$, $x \ne 0$

9. A student writes that the derivative of $x^2 \times x^3$ is $2x \times 3x^2 = 6x^3$

Summary

• The power rule: $f(x) = ax^n \Rightarrow f'(x) = anx^{n-1}$ for every integer $n$; the power comes down as a multiplier and drops by 1.

• The derivative of a constant is 0; the derivative of $mx$ is $m$.

• Constant multiples stay: $\frac{\mathrm{d}}{\mathrm{d}x}\big(af(x)\big) = af'(x)$

• The derivative of a polynomial of degree $n$ is a polynomial of degree $n - 1$.

• Terms such as $\frac{a}{x^k}$ are rewritten as $ax^{-k}$ first; their derivatives are $-kax^{-k-1}$

• Products are expanded and quotients with a single-term denominator are split into powers before differentiating; the derivative of a product is not the product of the derivatives.

• Letters other than $x$ work the same way ($\frac{\mathrm{d}V}{\mathrm{d}r}$, $\frac{\mathrm{d}s}{\mathrm{d}t}$

• The gradient at $x = a$ is $f'(a)$, a rate of change with units of output per unit of input; the GDC's numerical derivative finds it and checks a hand derivative.

Self-check

• To do:

  I can explain the pattern in the gradients of $x^2$, $x^3$ and $x^4$ that leads to the power rule.

• To do:

  I can differentiate $ax^n$ for positive, zero and negative integer $n$.

• To do:

  I can explain why the derivative of a constant is 0, using the graph of $y = c$.

• To do:

  I can differentiate a polynomial term by term and simplify every coefficient.

• To do:

  I can rewrite $\frac{3}{x}$, $\frac{5}{x^2}$

• To do:

  I can reduce a negative power by 1 correctly, so that $x^{-1}$ becomes $-x^{-2}$.

• To do:

  I can expand a product of brackets before differentiating.

• To do:

  I can split a quotient with a single-term denominator into separate powers of $x$.

• To do:

  I can differentiate with respect to $t$, $r$ or $q$, using the matching notation.

• To do:

  I can keep a parameter such as $k$ as a constant and then find it from a given gradient.

• To do:

  I can calculate $f'(a)$ by substituting the $x$-coordinate into the derivative.

• To do:

  I can interpret a derivative's value in context, with units such as $\text{cm}^3$ per cm.

• To do:

  I can find a numerical derivative on my GDC and use it to check a hand derivative.

• To do:

  I can solve an equation $f'(x) = m$ with my GDC and write down the equation I solved.

where $h$ is the height in metres at a horizontal distance $x$ metres from the start of the ramp, and $p$ and $q$ are constants. The two pieces meet at $x = 4$ with no jump and no sharp corner, so that $h$ is continuous and its gradient is continuous at $x = 4$.

1. (a)

   Find the value of $p$ and the value of $q$.

   [4]4 marks

   Markscheme (a)

   $h'(x) = 0.2x$ for $0 \le x \le 4$ (M1)

   $p = 0.2 \times 4 = 0.8$ A1

   $0.1 \times 4^2 = 4p + q$ (M1)

   $1.6 = 3.2 + q$ so $q = -1.6$ A1

   [4 marks]

2. (b)

   Find the horizontal distance from the start of the ramp at which the height of the ramp is 5 m.

   [2]2 marks

   Markscheme (b)

   $0.8x - 1.6 = 5$ (M1)

   $x = 8.25$ (m) A1

   Note: Follow through from part (a) provided $4 < x \le 10$

(c) $g'(x) = -\frac{2}{3} \times 6x^{5} = -4x^5$. Simplify the coefficient: $\frac{2}{3} \times 6 = 4$.

(d) $h'(x) = 4 \times 1 \times x^0 = 4$. The graph is a straight line with gradient 4.

(e) $k'(x) = 0$, because $-15$ is a constant.

(b) The gradient at $x = -1$ is $f'(-1)$. Substitute $x = -1$ into the derivative, with brackets round the negative number.

The gradient is $-32$: at $x = -1$ the graph is falling steeply, dropping about 32 units for each unit increase in $x$ close to that point.

Each coefficient is simplified: $\frac{4}{8} = \frac{1}{2}$ and $\frac{2}{3} \times 3 = 2$. Leaving $\frac{4x^3}{8}$ unsimplified is not a finished answer.

Differentiate term by term, taking care with the signs: the power $-1$ becomes $-2$, and $-2$ becomes $-3$.

(b) Substitute into the derivative.

The gradient is 2 at $x = 1$ and 2.125 at $x = 2$.

(b) From the derivative $g'(r) = -\frac{2GM}{r^3}$: