Differentiation: A-level Maths OCR revision notes

1.07

First principles and rules

The derivative from first principles 1.07

Definitions

Derivative: $f'(x) = \displaystyle\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}$ , the limit of the gradient of the chord joining $(x, f (x))$ to $(x + h, f (x + h))$ as the second point slides towards the first.
Gradient function: $\frac{d y}{d x}$ is itself a function of $x$ , giving the gradient of the tangent at each point of the curve $y = f (x)$ .

Method

Write down $f (x + h)$ and expand it completely.
Form the difference quotient $\frac{f ( x + h ) - f ( x )}{h}$ and simplify the numerator; every term not containing $h$ must cancel.
Divide the numerator through by $h$ .
Let $h \to 0$ , discarding every remaining term that still contains a factor of $h$ .

In practiceDifferentiate $f (x) = x^{3}$ from first principles.

$f (x + h) = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ expand fully
$\frac{f ( x + h ) - f ( x )}{h} = \frac{3 x ^{2} h + 3 x h ^{2} + h ^{3}}{h} = 3 x^{2} + 3 x h + h^{2}$ $x^{3}$ cancels; divide by $h$
$\displaystyle f'(x)=\lim_{h\to0}\left(3x^2+3xh+h^2\right)=3x^2$ the terms in $h$ vanish

Notes

First principles is rarely the efficient way to differentiate, but it is the definition that justifies every shortcut rule below, and questions sometimes demand it explicitly to test that understanding.
Keep the symbol $h \to 0 lim$ in front of the quotient until the final line; setting $h = 0$ before cancelling produces the meaningless $\frac{0}{0}$ .
For $f (x) = x^{2}$ , first principles gives $\frac{( x + h ) ^{2} - x ^{2}}{h} = \frac{2 x h + h ^{2}}{h} = 2 x + h \to 2 x$ — a concrete case worth rederiving from scratch to check the method is secure.

The power rule and index form 1.07

Key results

$\frac{d}{d x} (x^{n}) = n x^{n - 1}$ , for any real power $n$ .
$\frac{d}{d x} (k x^{n}) = kn x^{n - 1}$ , and the derivative of a constant is $0$ .
Derivatives add term by term: $\dfrac{d}{dx}\left(f(x)\pm g(x)\right) = f'(x)\pm g'(x)$ .
Negative powers: $\frac{1}{x ^{2}} = x^{- 2}$ differentiates to $- 2 x^{- 3} = - \frac{2}{x ^{3}}$ .
Fractional powers: $x = x^{\frac{1}{2}}$ differentiates to $\frac{1}{2} x^{- \frac{1}{2}} = \frac{1}{2 x}$ .
The tangent to $y = f (x)$ at $(x_{1}, y_{1})$ has gradient $m=f'(x_1)$ and equation $y - y_{1} = m (x - x_{1})$ ; the normal at the same point has gradient $- \frac{1}{m}$ .

Method

Rewrite every term as a power of $x$ first: expand brackets, split fractions with a single term in the denominator, and convert roots to fractional indices.
Differentiate term by term using the power rule.
Convert back into root or fraction form if the question was set in that form.

In practiceDifferentiate $y = \frac{x ^{3} + 4}{x}$ .

$y = x^{5/2} + 4 x^{- 1/2}$ split the fraction; $x = x^{1/2}$
$\frac{d y}{d x} = \frac{5}{2} x^{3/2} - 2 x^{- 3/2}$ power rule, term by term
$= \frac{5}{2} x x - \frac{2}{x x}$ back in root form

Notes

A quotient such as $\frac{3 x ^{2} + 1}{x}$ should be split into $3 x + x^{- 1}$ and differentiated directly; reaching for the quotient rule here wastes time and invites errors.
Worked illustration of a tangent: for $y = 2 x + \frac{3}{x} = 2 x^{\frac{1}{2}} + 3 x^{- 1}$ , the derivative is $x^{- \frac{1}{2}} - 3 x^{- 2}$ , which at $x = 1$ equals $1 - 3 = - 2$ ; since $y = 5$ there, the tangent is $y - 5 = - 2 (x - 1)$ , that is $y = 7 - 2 x$ .
The normal's gradient $- \frac{1}{m}$ is only defined when $m \neq = 0$ — at a stationary point, the tangent is horizontal ( $m = 0$ ) and the normal is vertical, with no gradient to quote.

Worked example

Differentiate:

f (x) = x^{2} - 3 x

from first principles, and hence state the gradient of the curve at $x = 4$ .

Show worked solution

f (x + h) = (x + h)^{2} - 3 (x + h) = x^{2} + 2 x h + h^{2} - 3 x - 3 h

Subtracting:

f (x) = x^{2} - 3 x

leaves $2 x h + h^{2} - 3 h$ , so:

\frac{f ( x + h ) - f ( x )}{h} = \frac{h ( 2 x + h - 3 )}{h} = 2 x + h - 3

Letting $h \to 0$ gives:

f&#039;(x)=2x-3

which agrees with the power rule.

At $x = 4$ the gradient is $2 (4) - 3 = 5$ .

1.07

Chain, product and quotient rules

The chain rule 1.07

Key results

If $y = f (g (x))$ then $\dfrac{dy}{dx} = f'(g(x))\cdot g'(x)$ .
Leibniz form: writing $u = g (x)$ gives $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ .
Linear inner function: $\frac{d}{d x} (a x + b)^{n} = an (a x + b)^{n - 1}$ .
Inverse form: $\frac{d y}{d x} = 1 \div \frac{d x}{d y}$ , useful when $x$ is given as a function of $y$ .

Method

Identify the inner function and call it $u$ .
Differentiate the outer function with respect to $u$ , keeping the inside unchanged.
Differentiate the inner function with respect to $x$ .
Multiply the two derivatives together and rewrite $u$ in terms of $x$ .

In practiceDifferentiate $y = (3 x^{2} - 1)^{5}$ .

$u = 3 x^{2} - 1, y = u^{5}$
$\frac{d y}{d u} = 5 u^{4}, \frac{d u}{d x} = 6 x$
$\frac{d y}{d x} = 5 u^{4} \cdot 6 x = 30 x (3 x^{2} - 1)^{4}$

Notes

The rule is often remembered as: differentiate the outside, keep the inside the same, then multiply by the derivative of the inside.
Forgetting the final multiplication by $g'(x)$ is the single most common differentiation slip; checking that the derivative of the inner function actually appears in the answer catches it every time.
Nested chains (a function inside a function inside a function) apply the rule repeatedly, multiplying together one derivative per layer — work from the outermost layer inwards, one at a time.

The product rule 1.07

Key results

If $y = uv$ then $\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$ .

Method

Split the expression into two factors and label them $u$ and $v$ .
Differentiate each factor separately, writing $\frac{d u}{d x}$ and $\frac{d v}{d x}$ down before combining anything.
Substitute into the formula.
Factorise the result, since a product-rule answer almost always has a common factor and a factorised form is what later parts of a question need.

In practiceDifferentiate $y = x^{2} e^{3 x}$ .

$u = x^{2}, \frac{d u}{d x} = 2 x; v = e^{3 x}, \frac{d v}{d x} = 3 e^{3 x}$ all four written first
$\frac{d y}{d x} = 2 x e^{3 x} + 3 x^{2} e^{3 x}$ $v \frac{d u}{d x} + u \frac{d v}{d x}$
$= x e^{3 x} (2 + 3 x)$ factorised: stationary at $x = 0$ and $x = - \frac{2}{3}$

Notes

Recognising when an expression is better rewritten before differentiating is part of the skill: expanding a product of two simple polynomials is quicker and safer than applying the product rule mechanically.
The product rule is not 'differentiate each factor and multiply' — $\frac{d}{d x} (uv) \neq = \frac{d u}{d x} \frac{d v}{d x}$ .
Writing down $u$ , $v$ , $\frac{d u}{d x}$ and $\frac{d v}{d x}$ explicitly before substituting into the formula avoids the common error of using the wrong pair mid-substitution.

The quotient rule 1.07

Key results

If $y = \frac{u}{v}$ then $\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ .

Notes

The order of the numerator matters, because subtraction is not commutative: the term beginning with $v$ comes first, and reversing it changes the sign of the whole answer.
The quotient rule is the product rule applied to $u v^{- 1}$ together with the chain rule, so either route is valid; the quotient rule is simply the tidier bookkeeping.
Choosing correctly between the chain, product and quotient rules — and spotting when none of them is needed — is often the real skill being tested rather than the differentiation itself.
As with the product rule, differentiate $u$ and $v$ separately and write both out before substituting — combining derivatives mentally mid-formula is where sign errors creep in.

Worked examples

Worked example

Differentiate:

y = (3 x^{2} + 1)^{4}

using the chain rule.

Show worked solution

Let the inside function be $u = 3 x^{2} + 1$ , so $y = u^{4}$ .

Then:

\frac{d y}{d u} = 4 u^{3}

and:

\frac{d u}{d x} = 6 x

By the chain rule:

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x} = 4 u^{3} \times 6 x = 24 x (3 x^{2} + 1)^{3}

Worked example

Differentiate:

y = \frac{sin x}{x ^{2}}

giving your answer as a single fraction in its simplest form.

Show worked solution

Use the quotient rule with $u = sin x$ , so:

\frac{d u}{d x} = cos x

and $v = x^{2}$ , so:

\frac{d v}{d x} = 2 x

Then:

\frac{d y}{d x} = \frac{x ^{2} cos x - 2 x sin x}{( x ^{2} ) ^{2}} = \frac{x ^{2} cos x - 2 x sin x}{x ^{4}}

Cancelling a factor of $x$ from numerator and denominator gives:

\frac{d y}{d x} = \frac{x cos x - 2 sin x}{x ^{3}}

valid for $x \neq = 0$ .

1.07

Differentiating trig, exp and log functions

Trigonometric, exponential and logarithmic derivatives 1.07

Key results

$\frac{d}{d x} (sin x) = cos x$ and $\frac{d}{d x} (cos x) = - sin x$ , with $x$ in radians.
$\frac{d}{d x} (e^{x}) = e^{x}$ , the defining property of $e$ from Exponentials and logarithms.
$\frac{d}{d x} (ln x) = \frac{1}{x}$ , for $x > 0$ .
Composite forms follow from the chain rule: $\frac{d}{d x} sin (a x + b) = a cos (a x + b)$ , $\dfrac{d}{dx}e^{f(x)} = f'(x)e^{f(x)}$ , and $\dfrac{d}{dx}\ln f(x) = \dfrac{f'(x)}{f(x)}$ .
For example, $\frac{d}{d x} (e^{3 x^{2}}) = 6 x e^{3 x^{2}}$ , since the inner function $3 x^{2}$ differentiates to $6 x$ .

Notes

These derivatives hold only in radians. In degrees the chain rule inserts an awkward factor, since $sin x^{\circ} = sin (\frac{π x}{180})$ , so calculus questions involving trigonometry are always set in radians.
The negative sign on the derivative of cosine is a frequent source of error and is worth checking deliberately in any answer containing it.
$ln f (x)$ 's derivative $\dfrac{f'(x)}{f(x)}$ is worth spotting directly in an integration context too — it's exactly the pattern that integrates back to $ln ∣ f (x) ∣$ .

1.07

Stationary points and curve sketching

Stationary points and their classification 1.07

Definitions

Stationary point: a point on the curve where $\frac{d y}{d x} = 0$ , so the tangent there is horizontal.
Point of inflection: a point where the curve changes its sense of curvature, that is, where $\frac{d ^{2} y}{d x ^{2}}$ changes sign.

Key results

$\frac{d ^{2} y}{d x ^{2}} > 0$ at a stationary point indicates a local minimum.
$\frac{d ^{2} y}{d x ^{2}} < 0$ at a stationary point indicates a local maximum.
$\frac{d ^{2} y}{d x ^{2}} = 0$ is inconclusive and requires checking the sign of $\frac{d y}{d x}$ on either side instead; this covers points of inflection as well as some maxima and minima the second-derivative test alone cannot distinguish.
Away from stationary points, $\frac{d y}{d x} > 0$ on an interval means the function is increasing there, and $\frac{d y}{d x} < 0$ means it is decreasing.

Method

Differentiate and solve $\frac{d y}{d x} = 0$ to find the $x$ -coordinates.
Substitute each root into the original equation to obtain the corresponding $y$ -coordinates.
Differentiate a second time and evaluate $\frac{d ^{2} y}{d x ^{2}}$ at each stationary point.
Classify using the sign of the second derivative, falling back on a sign test of $\frac{d y}{d x}$ either side of the point if the second derivative is zero.

In practiceFind and classify the stationary points of $y = x^{3} - 6 x^{2} + 9 x + 1$ .

$\frac{d y}{d x} = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3) = 0 ⟹ x = 1, 3$
$y (1) = 5, y (3) = 1$
$\frac{d ^{2} y}{d x ^{2}} = 6 x - 12 : - 6 < 0 at x = 1, 6 > 0 at x = 3$
$(1, 5) is a maximum, (3, 1) is a minimum$

Notes

The inconclusive case is genuinely ambiguous, not merely awkward: $y = x^{4}$ has $\frac{d ^{2} y}{d x ^{2}} = 0$ at $x = 0$ yet a clear minimum there, while $y = x^{3}$ has the same zero second derivative at $x = 0$ and a stationary point of inflection instead.
A stationary point is local, not global — the largest value of a function on a closed interval may occur at an endpoint rather than at any stationary point, so endpoints must be checked separately in optimisation problems.
Always substitute the $x$ -coordinate back into the ORIGINAL function, not the derivative, to find the $y$ -coordinate of a stationary point — a surprisingly common slip under time pressure.

Curve sketching 1.07

Method

Find where the curve crosses the axes, by setting $x = 0$ and then $y = 0$ .
Find the stationary points and classify them.
Determine the behaviour as $x \to \pm \infty$ , using the dominant term of the expression.
Identify any asymptotes, in particular values of $x$ excluded from the domain.
Draw a smooth curve consistent with all of the above, labelling every feature found.

In practiceSketch $y = \frac{x}{x - 2}$ .

$x = 0 ⟹ y = 0$ through the origin
$\frac{d y}{d x} = \frac{( x - 2 ) - x}{( x - 2 ) ^{2}} = \frac{- 2}{( x - 2 ) ^{2}} < 0$ no stationary points; decreasing on each branch
$y = 1 + \frac{2}{x - 2} \to 1 as x \to \pm \infty$ horizontal asymptote $y = 1$
$x \neq = 2$ vertical asymptote $x = 2$
Two branches: one through the origin to the left of $x = 2$ , below $y = 1$ ; the other above $y = 1$ to the right of $x = 2$ .

Notes

A sketch does not need to be to scale, but every feature shown on it should be justified by a calculation rather than guessed from a calculator display.
Sketching combines local information — stationary points and their nature — with global information about intercepts, end behaviour and asymptotes; a sketch missing either kind of information is incomplete.
For a rational function, check the behaviour on BOTH sides of any vertical asymptote separately — the curve can approach $+ \infty$ from one side and $- \infty$ from the other.

Worked examples

Worked example

Find and classify the stationary points of:

y = x^{3} - 3 x^{2} - 9 x + 5

Show worked solution

\frac{d y}{d x} = 3 x^{2} - 6 x - 9 = 3 (x^{2} - 2 x - 3) = 3 (x - 3) (x + 1)

Setting this to zero gives $x = 3$ or $x = - 1$ .

At $x = 3$ :

y = 27 - 27 - 27 + 5 = - 22

At $x = - 1$ :

y = - 1 - 3 + 9 + 5 = 10

The second derivative is:

\frac{d ^{2} y}{d x ^{2}} = 6 x - 6

At $x = 3$ :

6 (3) - 6 = 12 > 0

a local minimum at $(3, - 22)$ .

At $x = - 1$ :

6 (- 1) - 6 = - 12 < 0

a local maximum at $(- 1, 10)$ .

Worked example

The curve $y = x^{2} e^{3 x}$ has two stationary points.

Find their coordinates and determine the nature of each.

Show worked solution

Product rule with $u = x^{2}$ and $v = e^{3 x}$ :

\frac{d y}{d x} = 2 x e^{3 x} + x^{2} (3 e^{3 x}) = x e^{3 x} (2 + 3 x)

Since $e^{3 x} > 0$ always,

\frac{d y}{d x} = 0

requires $x = 0$ or:

x = - \frac{2}{3}

The $y$ -values are $y = 0$ at $x = 0$ , and:

y = \frac{4}{9} e^{- 2} = 0.0601

(3 s.f.) at:

x = - \frac{2}{3}

Differentiating:

\frac{d y}{d x} = (3 x^{2} + 2 x) e^{3 x}

again gives:

\frac{d ^{2} y}{d x ^{2}} = (6 x + 2) e^{3 x} + (3 x^{2} + 2 x) 3 e^{3 x} = (9 x^{2} + 12 x + 2) e^{3 x}

At $x = 0$ this is $2 > 0$ , a local minimum at $(0, 0)$ .

At:

x = - \frac{2}{3}

the bracket is $4 - 8 + 2 = - 2$ , so the second derivative is $- 2 e^{- 2} < 0$ , a local maximum at:

(- \frac{2}{3}, \frac{4}{9} e^{- 2})

1.07

Rates of change and connected rates

Rates of change and connected rates 1.07

Definitions

Rate of change: in context, $\frac{d y}{d x}$ measures how fast $y$ changes per unit change in $x$ , and carries the units of $y$ divided by the units of $x$ .

Key results

Connected rates follow from the chain rule: $\frac{d V}{d t} = \frac{d V}{d r} \times \frac{d r}{d t}$ , and hence $\frac{d r}{d t} = \frac{d V}{d t} \div \frac{d V}{d r}$ .
Any chain of linked variables works the same way, for example $\frac{d A}{d t} = \frac{d A}{d r} \times \frac{d r}{d t}$ for the surface area of an expanding sphere.

Method

Write down every rate given in the question in derivative notation, with its units.
Write down the formula linking the variables involved, such as $V = \frac{4}{3} π r^{3}$ .
Differentiate that formula to obtain the connecting derivative.
Combine the derivatives by the chain rule, then substitute the values that apply at the required instant.

In practiceA spherical balloon is inflated at 50 cm $^{3}$ s $^{- 1}$ . Find the rate at which the radius increases when $r = 10$ cm.

$\frac{d V}{d t} = 50, V = \frac{4}{3} π r^{3}$
$\frac{d V}{d r} = 4 π r^{2} = 400 π at r = 10$
$\frac{d r}{d t} = \frac{d V / d t}{d V / d r} = \frac{50}{400 π} = \frac{1}{8 π} \approx 0.0398 cm s^{- 1}$ chain rule

Notes

Substitute the specific value, such as $r = 10$ , only after differentiating; substituting first turns the formula into a constant whose derivative is zero.
Interpreting the answer matters as much as calculating it: state the units, and read a negative rate as a decrease rather than reporting it as an error.
Keep track of which variable each rate is 'per' — $\frac{d V}{d t}$ and $\frac{d V}{d r}$ look similar but mean very different things, and mixing them up is the most common error in connected-rates problems.

Worked example

A spherical balloon is inflated so that its volume increases at a constant rate of $50$ cm $^{3}$ s $^{- 1}$ .

Find the rate at which its radius is increasing at the instant when the radius is $10$ cm.

Show worked solution

The volume of a sphere is:

V = \frac{4}{3} π r^{3}

so:

\frac{d V}{d r} = 4 π r^{2}

At $r = 10$ this equals $400 π$ cm $^{2}$ .

By the chain rule,

\frac{d V}{d t} = \frac{d V}{d r} \times \frac{d r}{d t}

so:

\frac{d r}{d t} = \frac{d V}{d t} \div \frac{d V}{d r} = \frac{50}{400 π} = \frac{1}{8 π} = 0.0398

cm s

$^{- 1}$ (3 s.f.).

The value is positive, confirming the radius is increasing, and it is small because a large sphere needs a lot of extra volume for each extra centimetre of radius.