Lesson 4

Limits, and Differentiation

Limits, the gradient of a curve, differentiation from first principles and by rule, tangents and normals, stationary points and their nature, and the number e.

Taught 14 Sept 2026

Limits

The expression

\frac{(x + 1)(x - 2)}{x - 2}

is not defined when $x = 2$ , where it would read $\tfrac00$ . For every other value of $x$ the factor $x - 2$ cancels, and the expression is simply $x + 1$ . Its graph is the line $y = x + 1$ with one point missing: a hole at $x = 2$ .

Figure 4.1. The graph of

y = \dfrac{(x + 1)(x - 2)}{x - 2}

, which is the line

y = x + 1

with a hole at

(2, 3)

The expression is well defined for all values of $x$ other than $2$ , so we can legitimately ask what happens to it as $x$ approaches $2$ . Since it equals $x + 1$ at all those values, it approaches $3$ . We say that its limit as $x$ approaches $2$ is $3$ , and write

\lim_{x \to 2} \frac{(x + 1)(x - 2)}{x - 2} = 3 .

The value at $x = 2$ itself plays no part. The limit is about the values of the expression near $2$ , and here there is no value at $2$ at all.

Definition 4.1 (Limit).

Let $f(x)$ be defined for all values of $x$ near $a$ , except possibly at $a$ itself. We say that $f(x)$ tends to the limit $k$ as $x$ tends to $a$ , and write

\lim_{x \to a} f(x) = k,

if $f(x)$ gets and stays as close to $k$ as we want, provided $x$ is sufficiently close to $a$ .

It is not enough for the values to keep getting closer to $k$ . The numbers $2.1, 2.01, 2.001, \ldots$ get closer to $2$ , but each is also closer to $1$ than the one before, and we want to say that they approach $2$ and not $1$ . The difference is that they get as close to $2$ as we want, however small we set the allowed distance, whereas every one of them stays more than $1$ away from $1$ .

One-Sided Limits

Define $f(x)$ to be $x - 1$ when $x$ is negative and $x + 1$ when $x$ is positive. Its graph is two half-lines with a jump between them at $x = 0$ .

Figure 4.2. The graph of

f(x)

, equal to

x - 1

for negative

x

and to

x + 1

for positive

x

The limit of $f(x)$ as $x \to 0$ does not exist. We can make $x$ as close to $0$ as we want, but $f(x)$ does not get as close as we want to any one value: it is near $-1$ when $x$ is negative and near $1$ when $x$ is positive. We met the same difficulty with $\dfrac1x$ in the last lesson, where the behaviour depended on whether $x$ approached zero from above or from below; here both sides settle down, but to different values. We write

\lim_{x \to 0^+} f(x) = 1 \qquad\text{and}\qquad \lim_{x \to 0^-} f(x) = -1,

the first being the limit as $x$ tends to $0$ from above, or from the right, and the second the limit from below, or from the left. The limit $\displaystyle\lim_{x \to a} f(x)$ exists only when these two one-sided limits exist and are equal.

Limits at Infinity

We can also take limits as $x$ tends to infinity. If $g(x) = \dfrac1x$ , then $g(x)$ approaches $0$ as $x$ grows, and we write

\lim_{x \to \infty} g(x) = 0 .

As $x$ tends to minus infinity $g(x)$ also approaches $0$ , so $\displaystyle\lim_{x \to -\infty} g(x) = 0$ as well. For a limit at infinity, ” $x$ is sufficiently close to $a$ ” in the definition of a limit is replaced by ” $x$ is large enough”, and for minus infinity by ” $x$ is negative and large enough in size”.

Problem 4.1.

Find each limit, or show that it does not exist.

$\displaystyle\lim_{x \to 1} \frac{x^2 + x - 2}{x - 1}$ ;
$\displaystyle\lim_{x \to 1} f(x)$ , where $f(x) = x^2$ for $x < 1$ and $f(x) = 2x$ for $x > 1$ ;
$\displaystyle\lim_{x \to \infty} \frac{2x + 1}{x}$ .

The Gradient of a Curve

Chords, Tangents and Normals

Definition 4.2 (Chord, Tangent and Normal).

Let $A$ and $B$ be two points on a curve.

The line joining $A$ and $B$ is a chord of the curve.
The line touching the curve at $A$ is the tangent to the curve at $A$ .
The line through $A$ perpendicular to the tangent at $A$ is the normal to the curve at $A$ .

Figure 4.3. A chord

AB

, and the tangent and normal at

A

A straight line has the same gradient all along it. The gradient of a curve, which measures its slope, changes continually as we move along it. Suppose that as we move along the curve towards $A$ , the gradient stops changing at $A$ and stays constant from then on. We would then be moving along a straight line, and that line is the tangent at $A$ . So the gradient of the curve at $A$ is the same as the gradient of the tangent at $A$ .

Definition 4.3 (Gradient of a Curve).

The gradient of a curve at a point is the gradient of the tangent to the curve at that point. It measures the rate of increase of $y$ with respect to $x$ at that point.

If $B$ is another point on the curve, not too far from $A$ , the gradient of the chord $AB$ is an approximate value for the gradient of the tangent at $A$ , and the closer $B$ is to $A$ the better the approximation.

An approximate value can also be found by plotting the curve, drawing the tangent by eye and measuring its gradient. This method has to be used when we know the coordinates of a finite number of points on a curve but not its equation, as with the data from an experiment. When the equation of the curve is known we want an accurate method, so that we can take the analysis of curves and functions further.

Approaching the Tangent

Take the parabola $y = x^2$ and the point $(1.5, 2.25)$ on it. The tangent there, drawn on the left of Figure 4.4, seems to rise about $3$ units for each unit it moves to the right, so the gradient looks to be about $3$ . Is it exactly $3$ , and how would we find out?

Figure 4.4. The tangent to

y = x^2

(1.5, 2.25)

, and a close-up in which the dashed chord to

(1.6, 2.56)

runs almost along the tangent.

To approximate it, take a second point on the parabola close to the first, such as $(1.6, 2.56)$ , where $2.56 = 1.6^2$ because the point lies on $y = x^2$ . The line through $(1.5, 2.25)$ and $(1.6, 2.56)$ is found as in the last lesson. Its gradient is

\frac{2.56 - 2.25}{1.6 - 1.5} = \frac{0.31}{0.1} = 3.1,

and its equation is $y = 3.1x - 2.4$ . So our approximation to the gradient is $3.1$ , and the close-up in Figure 4.4 shows how near the chord runs to the tangent.

The approximation cannot be made perfect by taking both points to be $(1.5, 2.25)$ , since infinitely many lines pass through a single point and that tells us nothing. It can be made better. The chord through $(1.5, 2.25)$ and $(1.501, 2.253001)$ has gradient

\frac{2.253001 - 2.25}{1.501 - 1.5} = \frac{0.003001}{0.001} = 3.001,

so the gradients really do seem to be approaching $3$ . As $B$ approaches $A$ , the gradient of the chord $AB$ approaches the gradient of the tangent at $A$ , that is

\lim_{B \to A} \bigl(\text{gradient of chord } AB\bigr) = \text{gradient of tangent at } A .

Example 4.4.

Find the gradient of the curve $y = x(2x - 1)$ at the point $A$ where $x = 1$ .

When $x = 1$ , $y = 1$ , so $A$ is the point $(1, 1)$ . We calculate the gradient of a chord $AB$ and observe what happens to it as $B$ approaches $A$ . Take a succession of points $B_1, B_2, B_3, \ldots$ where $x = 1.5, 1.25, 1.125, \ldots$ , each time halving the remaining difference between the $x$ -coordinates of $A$ and $B$ .

$B$	$B_1$	$B_2$	$B_3$	$B_4$	$B_5$
$x$	$1.5$	$1.25$	$1.125$	$1.0625$	$1.03125$
gradient of $AB$	$4$	$3.5$	$3.25$	$3.125$	$3.0625$

As $B$ approaches $A$ the gradient of the chord $AB$ approaches $3$ , and we deduce that the gradient of the curve at $A$ is $3$ .

Figure 4.5. The chords

AB_1

AB_2

and

AB_3

y = x(2x - 1)

, turning towards the tangent at

A(1, 1)

The Delta Prefix

This numerical method is unsatisfactory, not least because of the amount of calculation involved, and in the end it only suggests the value $3$ . So instead of placing $B$ at particular positions we introduce a variable quantity for the difference between the $x$ -coordinates of $A$ and $B$ .

Definition 4.5 (Delta Prefix).

A variable quantity prefixed by $\delta$ means a small increase in that quantity. So $\delta x$ is a small increase in $x$ , and $\delta y$ is a small increase in $y$ .

Remark.

The letter $\delta$ is only a prefix. It cannot be treated as a factor: $\delta x$ is one quantity, not $\delta$ multiplied by $x$ .

Return to $y = x(2x - 1)$ and the point $A(1, 1)$ . If $\delta x$ is the increase in the $x$ -coordinate in moving from $A$ to $B$ , then the $x$ -coordinate of $B$ is $1 + \delta x$ . Every point on the curve satisfies $y = x(2x - 1)$ , so the $y$ -coordinate of $B$ is

(1 + \delta x)\bigl[2(1 + \delta x) - 1\bigr] = (1 + \delta x)(2\,\delta x + 1) .

The gradient of $AB$ is the increase in $y$ divided by the increase in $x$ :

\frac{(1 + \delta x)(2\,\delta x + 1) - 1}{\delta x} = \frac{3\,\delta x + 2(\delta x)^2}{\delta x} = 3 + 2\,\delta x .

As $B$ approaches $A$ , the difference between their $x$ -coordinates approaches zero, that is $\delta x \to 0$ . Therefore

\text{gradient at } A = \lim_{\delta x \to 0} \bigl(3 + 2\,\delta x\bigr) = 3 .

Dividing by $\delta x$ is allowed because a limit only involves values of $\delta x$ near zero, never zero itself.

The same calculation settles the question for the parabola. Writing $h$ for the small increase in $x$ from $1.5$ , the gradient of $y = x^2$ at $(1.5, 2.25)$ is

\lim_{h \to 0} \frac{(1.5 + h)^2 - 1.5^2}{h} = \lim_{h \to 0} \frac{1.5^2 + 3h + h^2 - 1.5^2}{h} = \lim_{h \to 0} \frac{3h + h^2}{h} = \lim_{h \to 0} \bigl(3 + h\bigr) = 3,

so it is exactly $3$ , as we guessed.

Example 4.6.

Find the gradient of the curve $y = \dfrac1x$ at the point where $x = 2$ .

Let $\delta x$ be the increase in $x$ in moving from $A\left(2, \tfrac12\right)$ to a nearby point $B$ on the curve, so that $B$ is $\left(2 + \delta x, \dfrac{1}{2 + \delta x}\right)$ . The gradient of the chord $AB$ is

\frac{1}{\delta x}\left(\frac{1}{2 + \delta x} - \frac12\right) = \frac{2 - (2 + \delta x)}{2(2 + \delta x)\,\delta x} = \frac{-\delta x}{2(2 + \delta x)\,\delta x} = -\frac{1}{2(2 + \delta x)} .

As $\delta x \to 0$ , $2(2 + \delta x) \to 4$ , so the gradient at $A$ is $-\tfrac14$ .

Problem 4.2.

Use the method of the last example to find the gradient of each curve at the point indicated.

$y = (x + 1)(x - 1)$ , where $x = 2$ ;
$y = 2x(x - 4)$ , where $x = 0$ ;
$y = (x + 2)(x - 1)$ , where $x = -3$ .

The Gradient Function

We found that the gradient of $y = x(2x - 1)$ is $3$ at the point where $x = 1$ . Instead of a fixed point, take $A$ to be any point $(x, y)$ on the curve. Its $y$ -coordinate can be written $x(2x - 1)$ , since both coordinates satisfy the equation of the curve. Let $B$ be another point on the curve such that the increase in the $x$ -coordinate in moving from $A$ to $B$ is $\delta x$ . Then $B$ is the point $\bigl[x + \delta x, (x + \delta x)(2x + 2\,\delta x - 1)\bigr]$ , and

\begin{aligned} \text{gradient of chord } AB &= \frac{(x + \delta x)(2x + 2\,\delta x - 1) - x(2x - 1)}{\delta x} \\ &= \frac{2x^2 + 4x\,\delta x + 2(\delta x)^2 - x - \delta x - 2x^2 + x}{\delta x} \\ &= \frac{4x\,\delta x - \delta x + 2(\delta x)^2}{\delta x} \\ &= 4x - 1 + 2\,\delta x . \end{aligned}

Then the gradient at any point $A$ on the curve is

\lim_{\delta x \to 0} \bigl(4x - 1 + 2\,\delta x\bigr) = 4x - 1 .

So the function $4x - 1$ gives the gradient at every point on the curve $y = x(2x - 1)$ . The gradient at a particular point, that is the rate of increase of $y$ with respect to $x$ there, is found by substituting the $x$ -coordinate of the point into $4x - 1$ . At $x = 1$ this gives $3$ , as before.

The function $4x - 1$ is called the gradient function of $y = x(2x - 1)$ , and the process of deriving it is called differentiation with respect to $x$ . Since $4x - 1$ was derived from $x(2x - 1)$ , it is also called the derivative, or derived function, of $x(2x - 1)$ . Using $\dfrac{d}{dx}$ as a symbol for “the derivative with respect to $x$ of”, we write

\frac{d}{dx}\bigl[x(2x - 1)\bigr] = 4x - 1,

or, when $y = x(2x - 1)$ ,

\frac{dy}{dx} = 4x - 1 .

Here $\dfrac{dy}{dx}$ means the derivative of $y$ with respect to $x$ , and is sometimes called the differential coefficient of $y$ . As an alternative notation we can use the symbol $D$ for “the derivative of”, so that $D\bigl[x(2x - 1)\bigr] = 4x - 1$ , and $D$ is referred to as the differential operator.

Since $4x - 1$ represents the rate of increase of $y$ with respect to $x$ , $\dfrac{dy}{dx}$ also represents the rate of increase of $y$ with respect to $x$ . Similarly $\dfrac{dv}{dt}$ means the derivative of $v$ with respect to $t$ , or the rate of increase of $v$ with respect to $t$ .

Example 4.7.

Differentiate $y = x^3 + 3$ with respect to $x$ .

Let $A$ be the point $(x, x^3 + 3)$ and $B$ the point $\bigl[x + \delta x, (x + \delta x)^3 + 3\bigr]$ on $y = x^3 + 3$ . Then

\text{gradient of } AB = \frac{(x + \delta x)^3 + 3 - (x^3 + 3)}{\delta x} = \frac{3x^2\,\delta x + 3x(\delta x)^2 + (\delta x)^3}{\delta x},

which simplifies to $3x^2 + 3x\,\delta x + (\delta x)^2$ . So the gradient at $A$ is $\displaystyle\lim_{\delta x \to 0} \bigl[3x^2 + 3x\,\delta x + (\delta x)^2\bigr] = 3x^2$ , that is

\frac{dy}{dx} = 3x^2 .

Problem 4.3.

Use the method of the last example to differentiate each of the following with respect to $x$ .

$y = x^2$ ;
$y = x^4$ ;
$y = 5x^2$ ;
$y = \dfrac1x$ .

Differentiating from First Principles

Consider any curve $y = f(x)$ . Let $A[x, f(x)]$ be any point on it, and let $\delta x$ be the increase in the $x$ -coordinate in moving from $A$ to another point $B$ on the same curve, so that $B$ is the point $[x + \delta x, f(x + \delta x)]$ . Let $\delta y$ be the corresponding increase in the $y$ -coordinate, that is

\delta y = f(x + \delta x) - f(x) .

Figure 4.6. The increases

\delta x

and

\delta y

in moving from

A

B

along

y = f(x)

The gradient of the chord $AB$ is

\frac{\delta y}{\delta x} = \frac{f(x + \delta x) - f(x)}{\delta x},

so the gradient at $A$ is

\frac{dy}{dx} = \lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} .

Definition 4.8 (Derivative).

Let $f$ be a function and $c$ a number. The derivative of $f$ at $c$ is the gradient of the graph of $f$ at the point where $x = c$ , defined by

\lim_{h \to 0} \frac{f(c + h) - f(c)}{h} .

A function for which this limit exists is differentiable at $c$ . The derivative of $f(x)$ is written $f'(x)$ or $\dfrac{d}{dx} f(x)$ , and when $y$ is a function of $x$ that we would graph, it is written $\dfrac{dy}{dx}$ .

The limit exists when the graph looks as though it has a slope at the point. If the graph had a spike there, it would not have a well-defined slope and the limit would not exist.

Using this definition to differentiate a function is called differentiating from first principles, and it can be used for any function, including functions we have not yet met. Fortunately it is not always necessary to go back to first principles, because whole families of functions can be differentiated by rules.

Remark.

We cannot tell how fast something is going from a single photograph of it. But we can photograph it at two nearby moments and divide the distance it moved by the time between the photographs, which gives a good approximation to its speed, and the closer the two moments, the better the approximation. This is how instruments that measure speed work, and it is the gradient of a chord standing in for the gradient of a tangent.

Rules of Differentiation

Differentiating a Constant

The equation $y = c$ , where $c$ is a constant, represents a straight line parallel to the $x$ -axis, so it has zero gradient:

\frac{d}{dx}(c) = 0 .

Differentiating $ax$

The equation $y = ax$ , where $a$ is a constant, represents a straight line with gradient $a$ , so

\frac{d}{dx}(ax) = a .

Differentiating $x^n$

The table collects results from the examples and problems above.

$f(x)$	$x$	$x^2$	$x^3$	$x^4$	$x^{-1}$
$\dfrac{d}{dx} f(x)$	$1$	$2x$	$3x^2$	$4x^3$	$-x^{-2}$

From this table it appears that to differentiate a power of $x$ we multiply by that power and then subtract one from the power, that is

\frac{d}{dx}\left(x^n\right) = n x^{n-1} . \tag{1}

This is the power rule.

The Power Rule for Whole Numbers

If we replace $1.5$ by $x$ and do the same calculation as for the parabola, the algebra gives

(x + h)^2 - x^2 = x^2 + 2xh + h^2 - x^2 = 2xh + h^2,

and after dividing by $h$ the remaining $h$ vanishes in the limit, so the parabola $y = x^2$ has gradient $2x$ at each value of $x$ . Expanding further powers shows a pattern:

\begin{aligned} (x + h)^2 - x^2 &= 2x\,h + h^2, \\ (x + h)^3 - x^3 &= 3x^2 h + 3x h^2 + h^3, \\ (x + h)^4 - x^4 &= 4x^3 h + 6x^2 h^2 + 4x h^3 + h^4 . \end{aligned}

In each line the term with $h$ to the first power is $n x^{n-1} h$ , and every other term contains $h^2$ or a higher power of $h$ . When $n$ is a positive whole number we show that this always happens by using the binomial expansion. The coefficients of $(x + h)^n$ are those of $(1 + x)^n$ , the numbers in row $n$ of Pascal’s triangle, so

(x + h)^n = x^n + (\text{second number in row } n)\, x^{n-1} h + (\text{terms containing } h^2, h^3, \ldots) .

The second number in a row is the sum of the two numbers above it, which are $1$ and the second number of the row before. So it goes up by $1$ from each row to the next, and as it is $1$ in row one, it is $n$ in row $n$ . Therefore

\frac{(x + h)^n - x^n}{h} = n x^{n-1} + (\text{terms containing } h, h^2, \ldots),

and as $h \to 0$ this tends to $n x^{n-1}$ .

The power rule, deduced here for whole numbers, is valid for all powers of $x$ , including fractional and negative powers, although this cannot be shown at this stage, and for the time being we take it on trust. Under the power conventions in Lesson 1, if $n$ is not a whole number then we use $x^n$ only where it is defined, and for the fractional powers considered here this means $x>0$ . The rule needs the power to be a fixed number: it does not find the gradient of $x^x$ , where the power changes with $x$ .

Example 4.9.

Accepting that the power rule can be used for any power of $x$ ,

\begin{aligned} \frac{d}{dx}\left(x^9\right) &= 9x^8, \\ \frac{d}{dx}\left(\sqrt{x}\right) &= \frac{d}{dx}\left(x^{1/2}\right) = \tfrac12 x^{-1/2} = \frac{1}{2\sqrt{x}} \quad (x>0), \\ \frac{d}{dx}\left(\frac{1}{x^3}\right) &= \frac{d}{dx}\left(x^{-3}\right) = -3x^{-4} = -\frac{3}{x^4} . \end{aligned}

Problem 4.4.

Differentiate each of the following with respect to $x$ by rule.

$x^7$ ;
$\dfrac{1}{x^2}$ ;
$\sqrt{x^3}$ ;
$\dfrac{1}{\sqrt[3]{x}}$ .

Differentiating $ax^n$

From the problems, $\dfrac{d}{dx}\left(5x^2\right) = 10x$ , which is $5$ times the derivative of $x^2$ . A constant factor always comes straight through, since

\frac{a(x + \delta x)^n - a x^n}{\delta x} = a \times \frac{(x + \delta x)^n - x^n}{\delta x},

and so

\frac{d}{dx}\left(a x^n\right) = a n x^{n-1}, \tag{2}

where $a$ is a constant.

Sums and Differences

From first principles,

\frac{(x + \delta x)^2 + 3(x + \delta x) - \left(x^2 + 3x\right)}{\delta x} = 2x + 3 + \delta x,

so $\dfrac{d}{dx}\left(x^2 + 3x\right) = 2x + 3$ , and in the same way $\dfrac{d}{dx}\left(x^2 - 2x + 1\right) = 2x - 2$ . Comparing these with the separate derivatives,

\begin{aligned} \frac{d}{dx}\left(x^2 + 3x\right) &= 2x + 3 = \frac{d}{dx}\left(x^2\right) + \frac{d}{dx}(3x), \\ \frac{d}{dx}\left(x^2 - 2x + 1\right) &= 2x - 2 = \frac{d}{dx}\left(x^2\right) - \frac{d}{dx}(2x) + \frac{d}{dx}(1) . \end{aligned}

So the operation “differentiate” is distributive across the addition and subtraction of functions, and gradients add:

\frac{d}{dx}\bigl[f(x) \pm g(x)\bigr] = \frac{d}{dx} f(x) \pm \frac{d}{dx} g(x) . \tag{3}

This holds in general, because the gradient of a chord of $y = f(x) + g(x)$ is the gradient of the corresponding chord of $y = f(x)$ plus that of $y = g(x)$ .

Example 4.10.

Using rules $(2)$ and $(3)$ ,

\begin{aligned} \frac{d}{dx}\left(3x^2 - 7x\right) &= 3(2x) - 7 = 6x - 7, \\ \frac{d}{dx}\left(x + \frac1x\right) &= \frac{d}{dx}\left(x + x^{-1}\right) = 1 - x^{-2} = 1 - \frac{1}{x^2}, \\ \frac{d}{dx}\left(x^3 - 2x^2\right) &= 3x^2 - 4x . \end{aligned}

Differentiation is not distributive across multiplication and division. For example

\frac{d}{dx}\bigl[x(x - 3)\bigr] = \frac{d}{dx}\left(x^2 - 3x\right) = 2x - 3, \qquad\text{but}\qquad \frac{d}{dx}(x) \times \frac{d}{dx}(x - 3) = 1 \times 1 = 1 .

Gradients do not multiply in the way they add. So in order to differentiate at this stage, any product must be expanded and any quotient must be divided out, to give terms which are added or subtracted, and no rule should be assumed that has not been established.

Example 4.11.

Differentiate $f(x) = \dfrac{4x^2 + x - 1}{2x}$ .

f(x) = \frac{4x^2}{2x} + \frac{x}{2x} - \frac{1}{2x} = 2x + \frac12 - \frac12 x^{-1},

therefore

f'(x) = 2 + 0 - \tfrac12\left(-x^{-2}\right) = 2 + \frac{1}{2x^2} .

Example 4.12.

Find the gradient of the curve $y = (x - 3)\left(x^2 + 2\right)$ at the point on the curve where $x = 1$ .

Expanding, $y = x^3 - 3x^2 + 2x - 6$ , therefore

\frac{dy}{dx} = 3x^2 - 6x + 2 .

When $x = 1$ , $\dfrac{dy}{dx} = 3 - 6 + 2 = -1$ , so the gradient of $y = (x - 3)\left(x^2 + 2\right)$ is $-1$ at the point where $x = 1$ .

Example 4.13.

Find the coordinates of the point on the curve $y = \dfrac{2}{x^2}$ at which its gradient is $\tfrac12$ .

Since $y = 2x^{-2}$ , $\dfrac{dy}{dx} = -4x^{-3}$ . The gradient is $\tfrac12$ when

-\frac{4}{x^3} = \frac12, \qquad\text{that is}\qquad x^3 = -8, \qquad x = -2 .

When $x = -2$ , $y = \dfrac{2}{4} = \tfrac12$ . Therefore the gradient of $y = \dfrac{2}{x^2}$ is $\tfrac12$ at the point $\left(-2, \tfrac12\right)$ .

Problem 4.5.

Differentiate $(x - 3)(2x + 5)$ and $\sqrt{x}\,(x - 1)$ with respect to $x$ .
Find the gradient of the curve $y = (2x - 3)(x + 1)$ at the point where $x = 0$ .
Find the coordinates of the point on the curve $y = \sqrt{x}$ at which the gradient is $2$ .

Tangents and Normals

The tangent to a curve at a point passes through that point and has the gradient of the curve there, so its equation comes from the line with a given gradient through a given point. When the tangent has non-zero gradient $m$ , the normal passes through the same point with the perpendicular gradient $-\dfrac1m$ . A horizontal tangent has a vertical normal.

Take $y = x^2$ at $(1.5, 2.25)$ once more. The tangent is the line of gradient $3$ through $(1.5, 2.25)$ , which is

y - 2.25 = 3(x - 1.5) .

Rearranging, $4y - 12x + 9 = 0$ is the equation of the tangent.

The normal to $y = x^2$ at $(1.5, 2.25)$ is the line through that point perpendicular to the tangent, so its gradient is $-\tfrac13$ and its equation is

y - 2.25 = -\tfrac13(x - 1.5), \qquad\text{that is}\qquad y + \tfrac13 x - \tfrac{11}{4} = 0 .

Multiplying through by $12$ to clear the fractions gives $12y + 4x - 33 = 0$ .

Figure 4.7. The parabola

y = x^2

with its tangent

4y - 12x + 9 = 0

and its normal

12y + 4x - 33 = 0

(1.5, 2.25)

Example 4.14.

Find the equation of the tangent to the curve $y = x^2 - 3x + 2$ at the point where it cuts the $y$ -axis.

The curve cuts the $y$ -axis where $x = 0$ , at $(0, 2)$ . The gradient of the tangent there is the value of $\dfrac{dy}{dx}$ when $x = 0$ . As $y = x^2 - 3x + 2$ , $\dfrac{dy}{dx} = 2x - 3$ , so when $x = 0$ the gradient of the curve is $-3$ .

Therefore the tangent has gradient $-3$ and passes through $(0, 2)$ . Its equation is $y = -3x + 2$ , that is

3x + y - 2 = 0 .

Example 4.15.

Find the equation of the normal to the curve $y = \dfrac1x$ at the point where $x = 2$ , and the coordinates of the point where this normal cuts the curve again.

As $y = x^{-1}$ , $\dfrac{dy}{dx} = -\dfrac{1}{x^2}$ . So when $x = 2$ , $y = \tfrac12$ and $\dfrac{dy}{dx} = -\tfrac14$ . Therefore the tangent at $\left(2, \tfrac12\right)$ has gradient $-\tfrac14$ , and the normal, the line perpendicular to the tangent, has gradient $4$ . The equation of the normal is

y - \tfrac12 = 4(x - 2), \qquad\text{that is}\qquad 8x - 2y - 15 = 0 .

The points of intersection of the normal and the curve are given by solving $8x - 2y - 15 = 0$ and $y = \dfrac1x$ simultaneously. Substituting for $y$ ,

8x - \frac2x - 15 = 0, \qquad\text{that is}\qquad 8x^2 - 15x - 2 = 0 .

We already know that the normal and the curve meet where $x = 2$ , so $x - 2$ is a factor, and

(x - 2)(8x + 1) = 0 .

Therefore they meet again where $x = -\tfrac18$ and $y = -8$ , at the point $\left(-\tfrac18, -8\right)$ .

Figure 4.8. The normal to

y = \dfrac1x

\left(2, \tfrac12\right)

meets the other branch of the curve at

\left(-\tfrac18, -8\right)

Problem 4.6.

Find the equations of the tangent and the normal to the curve $y = x^2 - 5x + 2$ at the point where $x = 3$ .
Find the equation of the tangent to $y = 2x^2 - 3x$ which has a gradient of $1$ .
Find the equation of the tangent to $y = (x - 5)(2x + 1)$ which is parallel to the $x$ -axis.

Stationary Values and Turning Points

Stationary Values

Definition 4.16 (Stationary Value).

A stationary value of a function $f(x)$ is a value of $f(x)$ at which its rate of change with respect to $x$ is zero, that is, where

\frac{d}{dx} f(x) = 0 .

A point on the curve $y = f(x)$ at which $\dfrac{dy}{dx} = 0$ is a stationary point.

At a stationary value of $f(x)$ the gradient of the curve $y = f(x)$ is zero, so the tangent to the curve is parallel to the $x$ -axis. The stationary values of $f(x)$ are the $y$ -coordinates of the points on $y = f(x)$ at which the tangent is parallel to the $x$ -axis.

The curve $y = x^3 - 2x^2$ in Figure 4.9 goes flat and turns round in two places, and the derivative tells us exactly where. We found that $\dfrac{d}{dx}\left(x^3 - 2x^2\right) = 3x^2 - 4x$ , and the graph is flat where

3x^2 - 4x = 0, \qquad\text{that is}\qquad x(3x - 4) = 0 .

The solutions are $x = 0$ and $x = \tfrac43$ , so it is exactly at these two values of $x$ that the graph is flat and turns round, at $(0, 0)$ and $\left(\tfrac43, -\tfrac{32}{27}\right)$ . A zero derivative does not always mean that the graph turns round, as we shall see with $y = x^3$ .

Figure 4.9. The curve

y = x^3 - 2x^2

, with horizontal tangents at

(0, 0)

and

\left(\tfrac43, -\tfrac{32}{27}\right)

Example 4.17.

Find the stationary values of $x^3 - 3x^2 + 2$ .

If $f(x) = x^3 - 3x^2 + 2$ , then $\dfrac{d}{dx} f(x) = 3x^2 - 6x$ . At stationary values of $f(x)$ this is zero, so

3x^2 - 6x = 0, \qquad 3x(x - 2) = 0 .

So the stationary values of $x^3 - 3x^2 + 2$ occur when $x = 0$ and when $x = 2$ . They are

f(0) = 2 \qquad\text{and}\qquad f(2) = 2^3 - 3\left(2^2\right) + 2 = -2 .

For the curve $y = x^3 - 3x^2 + 2$ , the gradient is zero at the points $(0, 2)$ and $(2, -2)$ .

The derivative also tells us which way a function is going. Where $\dfrac{dy}{dx}$ is positive the curve is rising, and we say that $y$ is increasing; where $\dfrac{dy}{dx}$ is negative the curve is falling, and $y$ is decreasing. For $y = x^3 - 2x^2$ the derivative $x(3x - 4)$ is positive for $x < 0$ , negative for $0 < x < \tfrac43$ and positive again for $x > \tfrac43$ , which is what Figure 4.9 shows.

Problem 4.7.

Find the values of $x$ at which $x^3 - 12x + 1$ has stationary values.
Find the stationary value of $3x^2 - 4x + 2$ .
Find the coordinates of the point on $y = (x - 2)(x + 3)$ at which the gradient is zero.

Turning Points

The gradient of a curve can be zero at several points. Close to one of these points, the shape of the curve belongs to one of the three kinds shown at $A$ , $B$ and $C$ in Figure 4.10.

Figure 4.10. A maximum turning point

A

, a minimum turning point

B

, and a point

C

where the gradient is zero but the curve does not turn.

Moving along the curve in the positive direction of the $x$ -axis:

near $A$ the gradient changes from positive, through zero at $A$ , to negative;
near $B$ the gradient changes from negative, through zero at $B$ , to positive;
at $C$ the gradient is zero, but it does not change sign as we move through $C$ , so the curve does not turn at $C$ . What does change at $C$ is the sense in which the curve is turning, from clockwise to anticlockwise.

Definition 4.18 (Turning Points and Points of Inflexion).

A point on a curve at which the gradient changes from positive, through zero, to negative is a maximum turning point. Its $y$ -coordinate is a maximum value of $y$ , or of $f(x)$ where $y = f(x)$ .
A point at which the gradient changes from negative, through zero, to positive is a minimum turning point. Its $y$ -coordinate is a minimum value of $y$ , or of $f(x)$ .
A point on a curve at which the sense of turning changes is a point of inflexion.

The gradient at a maximum or minimum turning point must be zero. The terms maximum value and minimum value do not mean the same as greatest value and least value: maxima and minima describe the behaviour of a function only in the immediate neighbourhood of its stationary values. In Figure 4.10, to the left of $A$ , the curve falls below the minimum value at $B$ .

Apart from $C$ there are two other points of inflexion in Figure 4.10, one between $A$ and $B$ and another between $B$ and $C$ , so the gradient at a point of inflexion is not necessarily zero. A stationary point can therefore be a maximum, a minimum, or neither. The simplest example of the last kind is $y = x^3$ at the origin: its gradient $3x^2$ is zero there, but the curve rises on both sides.

Figure 4.11. The curve

y = x^3

, stationary at the origin without turning.

The Nature of a Stationary Value

We know how to find the points on $y = f(x)$ at which $f(x)$ has stationary values, but to tell them apart we need to look further, and there are several ways of doing this. In Figure 4.10, let $A_1$ and $A_2$ be points on the curve close to $A$ and to its left and right respectively, and let $B_1$ , $B_2$ and $C_1$ , $C_2$ be placed in the same way about $B$ and $C$ .

Values of $y$

At $A$ , a maximum, $y$ at $A_1$ and $y$ at $A_2$ are both less than $y$ at $A$ . At $B$ , a minimum, $y$ at $B_1$ and $y$ at $B_2$ are both greater than $y$ at $B$ . At $C$ , a point of inflexion, $y$ at $C_1$ is less than $y$ at $C$ and $y$ at $C_2$ is greater.

	Maximum	Minimum	Inflexion
Values of $y$ either side of the stationary value	both smaller	both larger	one smaller and one larger

The Sign of the Gradient

At $A_1$ the gradient $\dfrac{dy}{dx}$ is positive, at $A$ it is zero and at $A_2$ it is negative. At $B_1$ it is negative, at $B$ zero and at $B_2$ positive. At $C_1$ it is positive, at $C$ zero and at $C_2$ positive again.

	Maximum	Minimum	Inflexion
Sign of $\dfrac{dy}{dx}$ moving through the stationary value	$+ \ 0 \ -$	$- \ 0 \ +$	$+ \ 0 \ +$ or $- \ 0 \ -$

The Second Derivative

When passing through $A$ , $\dfrac{dy}{dx}$ changes from positive to negative, so $\dfrac{dy}{dx}$ decreases as $x$ increases, that is, the rate of increase of $\dfrac{dy}{dx}$ with respect to $x$ is negative. The rate of increase of $\dfrac{dy}{dx}$ with respect to $x$ would be written $\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$ , and this clumsy notation is condensed to

\frac{d^2 y}{dx^2},

called the second derivative of $y$ with respect to $x$ . Similarly, when passing through $B$ , $\dfrac{dy}{dx}$ changes from negative to positive, so $\dfrac{dy}{dx}$ increases as $x$ increases and $\dfrac{d^2 y}{dx^2}$ is positive.

	Maximum	Minimum
Sign of $\dfrac{d^2 y}{dx^2}$	negative (or zero)	positive (or zero)

Points of inflexion are not so easily dealt with by this method. In the smooth examples considered here, $\dfrac{d^2 y}{dx^2} = 0$ at such points, but $\dfrac{d^2 y}{dx^2}$ can also be zero at maxima and minima.

The three tables summarise three alternative methods for determining the nature of a stationary value. The third method fails if $\dfrac{d^2 y}{dx^2}$ is found to be zero at the stationary value, and in that case one of the first two has to be used.

Example 4.19.

Find the points on $y = x^4 + 4x^3 - 6$ at which the gradient is zero, and determine the nature of these points.

y = x^4 + 4x^3 - 6, \qquad \frac{dy}{dx} = 4x^3 + 12x^2, \qquad \frac{d^2 y}{dx^2} = 12x^2 + 24x .

The gradient is zero when $4x^3 + 12x^2 = 0$ , that is $4x^2(x + 3) = 0$ , so when $x = -3$ and when $x = 0$ .

When $x = -3$ , $\dfrac{d^2 y}{dx^2} = 12(9) + 24(-3) = 36$ , which is positive, so $y$ has a minimum value here. It is $y = (-3)^4 + 4(-3)^3 - 6 = -33$ , so $(-3, -33)$ is a minimum turning point.

When $x = 0$ , $y = -6$ and $\dfrac{d^2 y}{dx^2} = 0$ , which is inconclusive. So we look at the sign of $\dfrac{dy}{dx} = 4x^2(x + 3)$ on either side of the point where $x = 0$ .

$x$	$-\tfrac12$	$0$	$\tfrac12$
$\dfrac{dy}{dx}$	$+$	$0$	$+$

The gradient does not change sign, so $(0, -6)$ is a point of inflexion.

Example 4.20.

Sketch the curve $y = 2x^3 + x^2 - 4x + 1$ .

Finding the maximum and minimum turning points gives a general idea of the shape and position of the curve.

y = 2x^3 + x^2 - 4x + 1, \qquad \frac{dy}{dx} = 6x^2 + 2x - 4, \qquad \frac{d^2 y}{dx^2} = 12x + 2 .

At turning points $6x^2 + 2x - 4 = 0$ , that is $2(3x - 2)(x + 1) = 0$ , so $x = \tfrac23$ or $x = -1$ .

When $x = \tfrac23$ , $\dfrac{d^2 y}{dx^2} = 12\left(\tfrac23\right) + 2 > 0$ and $y = 2\left(\tfrac23\right)^3 + \left(\tfrac23\right)^2 - 4\left(\tfrac23\right) + 1 = -\tfrac{17}{27}$ , so $\left(\tfrac23, -\tfrac{17}{27}\right)$ is a minimum turning point.
When $x = -1$ , $\dfrac{d^2 y}{dx^2} = 12(-1) + 2 < 0$ and $y = 2(-1)^3 + (-1)^2 - 4(-1) + 1 = 4$ , so $(-1, 4)$ is a maximum turning point.
As $y = 2x^3 + x^2 - 4x + 1$ , the curve cuts the $y$ -axis at $(0, 1)$ .

From these three results we can sketch the curve.

Figure 4.12. A sketch of

y = 2x^3 + x^2 - 4x + 1

from its turning points and its intercept on the

y

-axis.

In some cases the intercepts on the $x$ -axis can be found as well, although the equation which gives them is not always easy to solve. Here it is $2x^3 + x^2 - 4x + 1 = 0$ .

Example 4.21.

A farmer has an adjustable electric fence that is $100$ m long. It is used to enclose a rectangular grazing area on three sides, the fourth side being a fixed hedge. Find the maximum area that can be enclosed.

The length of the enclosure can be varied, up to $100$ m, and the width and area then depend on the length chosen. Let the length $SR$ be $x$ m. Then as $PS + SR + RQ = 100$ ,

PS = RQ = \tfrac12(100 - x),

and the area $A$ of the enclosure, in square metres, is

A = x\left[\tfrac12(100 - x)\right] = 50x - \tfrac12 x^2 .

Now $A$ is a quadratic function of $x$ with a negative coefficient of $x^2$ , so from what we know of quadratic functions it is greatest at $x = 50$ , halfway between its zeros at $x = 0$ and $x = 100$ . So the maximum area that can be enclosed is $50 \times 25 = 1250$ square metres.

The maximum value of $50x - \tfrac12 x^2$ can also be found by differentiation. We have

\frac{dA}{dx} = 50 - x,

and $A$ has a stationary value when $\dfrac{dA}{dx} = 0$ , that is when $x = 50$ . Now $\dfrac{d^2 A}{dx^2} = -1$ , which is negative, so this is a maximum, and the maximum value is $25 \times 50 = 1250$ square metres.

Figure 4.13. The enclosure against the hedge, and its area

A = 50x - \tfrac12 x^2

as the length

x

varies.

The second method, using differentiation, is necessary when finding maximum or minimum values of functions that are not quadratic. For quadratic functions the first method is preferable, since their properties let us find the maximum or minimum value by inspection. In this case the greatest value and the maximum value are the same.

Example 4.22.

A rectangle has a perimeter of $20$ units. Find its greatest possible area.

The perimeter, the total length of the edges, is twice the short side plus twice the long side, so the two sides add up to $10$ . So if one side has length $x$ , the other has length $10 - x$ , and the area of the rectangle is

x(10 - x) = 10x - x^2 .

This is differentiable everywhere, with derivative $10 - 2x$ . At its greatest value, either its gradient is $0$ or $x$ is at one of the ends of its range, $x = 0$ or $x = 10$ . At the ends the area is $0$ , so we solve $10 - 2x = 0$ , which gives $x = 5$ . The area there is $10(5) - 5^2 = 25$ , so the greatest area is $25$ square units.

The point where the gradient is zero is a maximum and not a minimum, for three separate reasons.

$x = 5$ is the only place where $10x - x^2$ can turn round, and at values of $x$ on either side of it, such as $0$ and $10$ , the area is less than $25$ . If the area were more than $25$ at $x = 6$ , say, the curve would have to turn round again to reach $(10, 0)$ , and it does not.
The gradient $10 - 2x$ is positive to the left of $x = 5$ and negative to the right.
The graph of $10x - x^2$ is a parabola opening downwards, so its only flat point is a maximum.

Remark.

If $f(x) = ax^2 + bx + c$ , then $f'(x) = 2ax + b$ , so $f(x)$ has a stationary value where $x = -\dfrac{b}{2a}$ , on the axis of the curve. If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$ , where the curve crosses the $x$ -axis, then by the sum of the roots $\tfrac12(\alpha + \beta) = -\dfrac{b}{2a}$ , so the turning point of the curve has $x$ -coordinate $\tfrac12(\alpha + \beta)$ , halfway between the roots. A ball thrown through the air follows such a parabola, if we ignore air resistance.

Problem 4.8.

Find the stationary values of each of the following functions and determine their nature.

$2x^3 - 3x^2 - 12x$ ;
$(x - 3)(2x + 1)$ ;
$x^3 + 3$ .

The Number $e$

Exponential Growth

The curve $y = 2^x$ of the last lesson increases as $x$ increases. Its values at $x = -3, -2, -1, 0, 1, 2, 3$ are $\tfrac18, \tfrac14, \tfrac12, 1, 2, 4, 8$ , and the graph grows quite fast. Differentiating from first principles shows how fast:

\frac{2^{x + h} - 2^x}{h} = \frac{2^x \cdot 2^h - 2^x}{h} = 2^x \cdot \frac{2^h - 1}{h},

and the factor $\dfrac{2^h - 1}{h}$ does not involve $x$ at all. Its values, and those of the corresponding factor for $3^x$ , as $h$ approaches zero are

$h$	$0.1$	$0.01$	$0.001$	$0.0001$
$\dfrac{2^h - 1}{h}$	$0.7177$	$0.6956$	$0.6934$	$0.6932$
$\dfrac{3^h - 1}{h}$	$1.1612$	$1.1047$	$1.0992$	$1.0987$

so the gradient of $2^x$ is about $0.693 \times 2^x$ , and the gradient of $3^x$ is about $1.099 \times 3^x$ .

The rate of change of an exponential function is proportional to its current value. The larger the function gets, the faster it grows, which makes it larger still; this runaway growth is called exponential growth. It is what happens with the spread of a disease, or with interest in a bank.

A base for which the factor is exactly $1$ would give a function whose derivative is equal to the function itself. The factor is less than $1$ for base $2$ and greater than $1$ for base $3$ , and there is a number between $2$ and $3$ for which it is exactly $1$ .

Definition 4.23 (The Number e).

The number $e$ , which is about $2.71828182845904523536\ldots$ , is the base for which

\frac{d}{dx}\left(e^x\right) = e^x .

With $e$ in place of $2$ the factor takes the values $1.0517$ , $1.0050$ , $1.0005$ and $1.0001$ for the four values of $h$ in the table. That such a number exists, and that $e^x$ has exactly this derivative, we accept on trust at this stage, as we did the power rule for fractional powers.

Figure 4.14. The curve

y = e^x

. At every point the gradient equals the height: the tangent at

(0, 1)

has gradient

1

, and the tangent at

(1, e)

has gradient

e

Example 4.24.

Find the equation of the tangent to $y = e^x$ at the point $(0, 1)$ .

Since $\dfrac{dy}{dx} = e^x$ , the gradient at $(0, 1)$ is $e^0 = 1$ . The tangent is the line of gradient $1$ through $(0, 1)$ , which is $y = x + 1$ .

The Natural Logarithm

Definition 4.25 (Natural Logarithm).

The natural logarithm of a positive number $x$ is its logarithm to base $e$ , written

\ln x = \log_e x .

So $\ln x$ is the number $y$ for which $e^y = x$ . Since $\log_a 1 = 0$ and $\log_a\left(a^k\right) = k$ for every base, straight from the definition of a logarithm, we have $\ln 1 = 0$ , $\ln e = 1$ and $\ln\left(e^k\right) = k$ .

As a function from the positive numbers to the real numbers, $\ln x$ is the inverse of the function $e^x$ from the real numbers to the positive numbers, in the same way that $\log_2 x$ is the inverse of $2^x$ . Its graph is the reflection of $y = e^x$ in the line $y = x$ , and has the same shape as the graph of $y = \log_2 x$ .

Remark.

People often write $\log x$ with no base. If you are a physicist or an engineer this usually means $\log_{10} x$ , but mathematicians usually mean $\ln x$ . In these notes base $10$ is written $\lg$ , base $e$ is written $\ln$ , and $\log$ without a base appears only in rules that hold for every base.

Example 4.26.

Find the coordinates of the stationary point of $y = e^x - 3x$ , and determine its nature.

\frac{dy}{dx} = e^x - 3, \qquad \frac{d^2 y}{dx^2} = e^x .

The gradient is zero when $e^x = 3$ , that is when $x = \ln 3$ , and there $y = e^{\ln 3} - 3\ln 3 = 3 - 3\ln 3$ . Since $e^x$ is positive for every $x$ , $\dfrac{d^2 y}{dx^2} > 0$ , so $(\ln 3, 3 - 3\ln 3)$ , which is about $(1.099, -0.296)$ , is a minimum turning point.

Problem 4.9.

Solve the equation $e^x = 10$ , giving $x$ exactly and to three decimal places.
Differentiate $x^3 + e^x$ with respect to $x$ .
Find the gradient of the curve $y = e^x - x^2$ at the point where $x = 0$ .

Exercises

Exercise 4.1.

Differentiate $f(x) = x + \dfrac1x$ from first principles.

Exercise 4.2.

Find the equations of the tangents to the curve $y = (2x - 1)(x + 1)$ at the points where the curve cuts the $x$ -axis, and find the point of intersection of these tangents.

Exercise 4.3.

Find the coordinates of the point on $y = x^2 - 5$ at which the gradient is $3$ . Hence find the value of $c$ for which the line $y = 3x + c$ is a tangent to $y = x^2 - 5$ .

Exercise 4.4.

Find the value of $k$ for which $y = 2x + k$ is a normal to $y = 2x^2 - 3$ .

Exercise 4.5.

Find the coordinates of the point on $y = x^2 - 7x + 3$ at which the gradient is $2$ . Hence find the equation of the normal to $y = x^2 - 7x + 3$ which is parallel to $x + 2y - 1 = 0$ .

Exercise 4.6.

Show that $\left(t, \dfrac1t\right)$ lies on the curve $y = \dfrac1x$ for all non-zero values of $t$ . Find the equation of the tangent to $y = \dfrac1x$ at $\left(t, \dfrac1t\right)$ , and the area of the triangle enclosed by this tangent and the coordinate axes.

Exercise 4.7.

Find the turning points on $y = 3x^4 + 4x^3 - 12x^2$ , and give a rough sketch of the curve.

Exercise 4.8.

Find the stationary values of $f(x) = x + \dfrac1x$ , and use them to sketch the curve $y = x + \dfrac1x$ .

Exercise 4.9.

Find the coordinates of the turning points of the curve $y = 3 + 24x - 21x^2 - 4x^3$ , and sketch the curve.

Exercise 4.10.

The gradient function of $y = ax^2 + bx + c$ is $4x + 2$ , and the function has a minimum value of $1$ . Find the values of $a$ , $b$ and $c$ .

Exercise 4.11.

A closed cylindrical can has height $h$ and base radius $r$ , and its volume is $0.01$ m³. Show that

h = \frac{1}{100\pi r^2} .

Show further that $S$ , the surface area, is given by

S = 2\pi r^2 + \frac{1}{50r},

and hence find the value of $r$ for which $S$ is a minimum.

Exercise 4.12.

An open rectangular box is made from a square sheet of cardboard by removing a square from each corner and joining the cut edges. If the cardboard has edge $0.5$ m, find the maximum volume of the box.

Exercise 4.13.

A cylinder is cut from a solid sphere of radius $5$ cm, so that the curved edges of the cylinder reach the surface of the sphere. If the height of the cylinder is $2h$ , show that its volume is $2\pi h\left(25 - h^2\right)$ , and find the maximum volume of such a cylinder.

Exercise 4.14.

A piece of string of fixed length is made to enclose a rectangle. Show that the enclosed area is greatest when the rectangle is a square.

Exercise 4.15.

The table shows the results of an experiment in which some hot liquid was left to cool, its temperature $\theta$ °C being measured at intervals of one minute.

Time $t$ (minutes)	$0$	$1$	$2$	$3$	$4$	$5$
Temperature $\theta$ (°C)	$100$	$93$	$87$	$83$	$80$	$78$

What does $\dfrac{d\theta}{dt}$ represent? By drawing a graph of these results, estimate the rate of decrease of the temperature after three minutes.

Exercise 4.16.

Show that the tangent to $y = e^x$ at the point where $x = 1$ passes through the origin. Hence find the value of $k$ for which the line $y = kx$ is a tangent to $y = e^x$ .

Check Yourself

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Limits

One-Sided Limits

Limits at Infinity

The Gradient of a Curve

Chords, Tangents and Normals

Approaching the Tangent

The Delta Prefix

The Gradient Function

Differentiating from First Principles

Rules of Differentiation

Differentiating a Constant

Differentiating axaxax

Differentiating xnx^nxn

The Power Rule for Whole Numbers

Differentiating axnax^naxn

Sums and Differences

Tangents and Normals

Stationary Values and Turning Points

Stationary Values

Turning Points

The Nature of a Stationary Value

Values of yyy

The Sign of the Gradient

The Second Derivative

The Number eee

Exponential Growth

The Natural Logarithm

Exercises

Check Yourself

Differentiating $ax$

Differentiating $x^n$

Differentiating $ax^n$

Values of $y$

The Number $e$