1.07PureCore

Derivatives of Functions

Q: What does dy dx actually mean?

The rate of change of y with respect to x — the gradient of the curve at a point. It is a single symbol, not a fraction, although it often behaves like one.

Q: Can I avoid the quotient rule?

Yes: write uv as uv^-1 and use the product and chain rules. The result is the same, though it may need simplifying.

Q: Why must trig derivatives use radians?

The limit hh 1 only holds in radians. In degrees every derivative gains a factor of 180.

The gradient of a straight line is the same everywhere. The gradient of a curve changes from point to point, and $differentiation$ finds a formula for it. That single idea — the rate of change at an instant — lets you find the steepness of a curve, the velocity of a moving object, and the maximum and minimum values of anything you can write as a function.

50 min Differentiation

The big picture

The derivative is defined as a limit: the gradient of a chord between two points on the curve, as the points slide together. Working that limit out once for $x^{n}$ gives the power rule, and everything else in the topic is built from a short list of standard derivatives plus three combination rules. The chain rule handles functions inside functions; the product rule handles functions multiplied together; the quotient rule handles one divided by another. Learn to recognise the structure of an expression — which is exactly the decomposition skill from the functions chapter — and differentiating even complicated expressions becomes mechanical.

What you'll be able to do

Understand the derivative as the limit of the gradient of a chord, and differentiate from first principles
Differentiate $x^{n}$ , $e^{k x}$ , $a^{x}$ , $ln x$ , $sin k x$ , $cos k x$ and $tan k x$
Apply the chain rule, product rule and quotient rule
Use Leibniz notation $\frac{d y}{d x}$ and function notation $f^{'} (x)$

Differentiation from first principles

The chord from $(x, f (x))$ to $(x + h, f (x + h))$ has gradient $\frac{f ( x + h ) - f ( x )}{h}$ . As $h \to 0$ , the chord approaches the tangent, and its gradient approaches the $derivative$ .

OCR expects you to carry out this limit for simple polynomials, and to show that $h$ is cancelled $before$ it is set to zero.

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

The definition of the derivative.

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

2Divide by

h

3 x^{2} + 3 x h + h^{2}

3As

h \to 0

, the terms

3 x h

and

h^{2}

tend to

0

Answer

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

Why this works

You cannot simply put $h = 0$ into the chord gradient — that gives $\frac{0}{0}$ . The algebra that cancels $h$ first is the whole point of the method, and it is what the marks are for.

Standard derivatives

$Powers$ : $\frac{d}{d x} x^{n} = n x^{n - 1}$ for any rational $n$ . Rewrite roots and reciprocals as powers first: $\frac{1}{x} = x^{- 1/2}$ .

$Exponentials and logs$ : $e^{k x}$ differentiates to $k e^{k x}$ ; $a^{x}$ to $a^{x} ln a$ ; and $ln x$ to $\frac{1}{x}$ .

$Trigonometric functions$ (in radians): $sin k x \to k cos k x$ , $cos k x \to - k sin k x$ , $tan k x \to k sec^{2} k x$ .

Derivatives of sums are sums of derivatives, and constant multiples carry through.

\frac{d}{d x} x^{n} = n x^{n - 1}

Power rule.

\frac{d}{d x} e^{k x} = k e^{k x}, \frac{d}{d x} ln x = \frac{1}{x}

Exponential and log.

\frac{d}{d x} sin k x = k cos k x, \frac{d}{d x} cos k x = - k sin k x

Trigonometric, radians only.

1Rewrite:

y = 4 x^{1/2} - 3 x^{- 2} + 5 e^{2 x}

2Differentiate term by term.

Answer

\frac{d y}{d x} = 2 x^{- 1/2} + 6 x^{- 3} + 10 e^{2 x}

Tip — The trigonometric derivatives only hold in radians. In degrees an extra factor of $\frac{π}{180}$ appears, which is why calculus always uses radians.

The chain rule

For a composite $y = f (g (x))$ , let $u = g (x)$ . Then $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ .

In words: differentiate the outer function, leaving the inside alone, then multiply by the derivative of the inside.

A related result: $\frac{d y}{d x} = \frac{1}{d x / d y}$ , useful when $x$ is given as a function of $y$ .

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}

Chain rule.

1First: outer

u^{5}

gives

5 u^{4}

; inner derivative

6 x

. So

5 (3 x^{2} - 1)^{4} \times 6 x

2Second: outer

ln u

gives

\frac{1}{u}

; inner derivative

- sin x

. So

\frac{- sin x}{cos x}

Answer

30 x (3 x^{2} - 1)^{4}

and

- tan x

Product and quotient rules

For a product $y = uv$ of two functions of $x$ : $\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$ .

For a quotient $y = \frac{u}{v}$ : $\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ . The order in the numerator matters because of the minus sign.

Set out $u$ , $v$ , $u^{'}$ and $v^{'}$ as four separate lines before combining them. It keeps the working checkable.

\frac{d}{d x} (uv) = u v^{'} + v u^{'}

Product rule.

\frac{d}{d x} (\frac{u}{v}) = \frac{v u ^{'} - u v ^{'}}{v ^{2}}

Quotient rule.

1Product:

u = x^{2}

u^{'} = 2 x

;

v = e^{3 x}

v^{'} = 3 e^{3 x}

. So

3 x^{2} e^{3 x} + 2 x e^{3 x}

2Quotient:

u = sin x

u^{'} = cos x

;

v = x

v^{'} = 1

. So

\frac{x cos x - sin x}{x ^{2}}

Answer

x e^{3 x} (3 x + 2)

and

\frac{x cos x - sin x}{x ^{2}}

Tip — Factorise the answer to a product-rule derivative. Setting it equal to zero, as you will for stationary points, is far easier in factorised form.

Think like an examiner

In first-principles proofs, cancel $h$ before letting $h \to 0$ , and write the limit notation.
Rewrite roots and reciprocals as powers before differentiating.
Show $u$ , $v$ , $u^{'}$ , $v^{'}$ explicitly for product and quotient rules.
Work in radians whenever trigonometric functions are differentiated.

Common misconceptions

Differentiating

e^{3 x}

3 x e^{3 x - 1}

Exponentials are not powers of

x

\frac{d}{d x} e^{3 x} = 3 e^{3 x}

Differentiating

(2 x + 1)^{4}

4 (2 x + 1)^{3}

Multiply by the inner derivative:

8 (2 x + 1)^{3}

Writing

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

\frac{d}{d x} cos x = - sin x

Differentiating a product as the product of the derivatives.

Use

u v^{'} + v u^{'}

Differentiation rules

\frac{d}{d x} x^{n} = n x^{n - 1}

Power rule.

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

Chain rule.

(uv)^{'} = u v^{'} + v u^{'}

Product rule.

(\frac{u}{v})^{'} = \frac{v u ^{'} - u v ^{'}}{v ^{2}}

Quotient rule.

Stretch yourself

Prove from first principles that the derivative of $sin x$ is $cos x$ , given that $\frac{sin h}{h} \to 1$ and $\frac{cos h - 1}{h} \to 0$ as $h \to 0$ (with $h$ in radians).

Hint — Expand $sin (x + h)$ with the compound angle formula, then group the terms containing $sin x$ and $cos x$ .

Questions students ask

What does

\frac{d y}{d x}

actually mean?

The rate of change of

y

with respect to

x

— the gradient of the curve at a point. It is a single symbol, not a fraction, although it often behaves like one.

When do I use the chain rule?

Whenever a function is applied to something other than plain

x

— a bracket to a power,

e

to a power, sine of an expression, and so on.

Can I avoid the quotient rule?

Yes: write

\frac{u}{v}

u v^{- 1}

and use the product and chain rules. The result is the same, though it may need simplifying.

Why must trig derivatives use radians?

The limit

\frac{sin h}{h} \to 1

only holds in radians. In degrees every derivative gains a factor of

\frac{π}{180}

Key takeaways

The derivative is the limit of the chord gradient as $h \to 0$ .
Power rule $n x^{n - 1}$ ; $e^{k x} \to k e^{k x}$ ; $ln x \to \frac{1}{x}$ ; trig derivatives in radians.
Chain rule for composites: outer derivative times inner derivative.
Product rule $u v^{'} + v u^{'}$ ; quotient rule $\frac{v u ^{'} - u v ^{'}}{v ^{2}}$ .
Rewrite expressions as powers and factorise results.

How this fits the course

Build on

Composite Functions

Exponential Functions

Leads to

Tangents and Normals

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