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 $\textbf{differentiation}$ 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 $x^{n}$ , $\mathrm{e}^{kx}$ , $a^{x}$ , $\ln x$ , $\sin kx$ , $\cos kx$ and $\tan kx$
Apply the chain rule, product rule and quotient rule
Use Leibniz notation $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ and function notation $\mathrm{f}^{\prime}(x)$

Differentiation from first principles

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

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

\mathrm{f}^{\prime}(x) = \lim_{h \to 0} \dfrac{\mathrm{f}(x + h) - \mathrm{f}(x)}{h}

The definition of the derivative.

Example 1Differentiate

\mathrm{f}(x) = x^{3}

from first principles.

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

2Divide by

h

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

3As

h \to 0

, the terms

3xh

and

h^{2}

tend to

0

Answer

\mathrm{f}^{\prime}(x) = 3x^{2}

Why this works

You cannot simply put $h = 0$ into the chord gradient — that gives $\tfrac00$ . The algebra that cancels $h$ first is the whole point of the method, and it is what the marks are for.

Standard derivatives

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

$\textbf{Exponentials and logs}$ : $\mathrm{e}^{kx}$ differentiates to $k\mathrm{e}^{kx}$ ; $a^{x}$ to $a^{x}\ln a$ ; and $\ln x$ to $\dfrac{1}{x}$ .

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

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

\dfrac{\mathrm{d}}{\mathrm{d}x}x^{n} = nx^{n-1}

Power rule.

\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{kx} = k\mathrm{e}^{kx}, \qquad \dfrac{\mathrm{d}}{\mathrm{d}x}\ln x = \dfrac{1}{x}

Exponential and log.

\dfrac{\mathrm{d}}{\mathrm{d}x}\sin kx = k\cos kx, \qquad \dfrac{\mathrm{d}}{\mathrm{d}x}\cos kx = -k\sin kx

Trigonometric, radians only.

Example 1Differentiate

y = 4\sqrt{x} - \dfrac{3}{x^{2}} + 5\mathrm{e}^{2x}

1Rewrite:

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

2Differentiate term by term.

Answer

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

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

The chain rule

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

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

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

\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u} \times \dfrac{\mathrm{d}u}{\mathrm{d}x}

Chain rule.

Example 1Differentiate

y = (3x^{2} - 1)^{5}

and

y = \ln(\cos x)

1First: outer

u^{5}

gives

5u^{4}

; inner derivative

6x

. So

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

2Second: outer

\ln u

gives

\dfrac{1}{u}

; inner derivative

-\sin x

. So

\dfrac{-\sin x}{\cos x}

Answer

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

and

-\tan x

Product and quotient rules

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

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

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

\dfrac{\mathrm{d}}{\mathrm{d}x}(uv) = uv^{\prime} + vu^{\prime}

Product rule.

\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{u}{v}\right) = \dfrac{vu^{\prime} - uv^{\prime}}{v^{2}}

Quotient rule.

Example 1Differentiate

y = x^{2}\mathrm{e}^{3x}

and

y = \dfrac{\sin x}{x}

1Product:

u = x^{2}

u^{\prime} = 2x

;

v = \mathrm{e}^{3x}

v^{\prime} = 3\mathrm{e}^{3x}

. So

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

2Quotient:

u = \sin x

u^{\prime} = \cos x

;

v = x

v^{\prime} = 1

. So

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

Answer

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

and

\dfrac{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.

Prove both trigonometric derivatives from first principles

Use the radian limits $\sin h/h\to1$ and $(\cos h-1)/h\to0$ . The compound-angle formula separates the h-dependent terms so those limits can be used. This is a derivation, rather than merely quoting a derivative rule.

For sine, the difference quotient becomes $\sin x\,\dfrac{\cos h-1}{h}+\cos x\,\dfrac{\sin h}{h}$ , whose limit is $\cos x$ . The cosine proof below uses the same two limits.

Example 1Derive

d(\cos x)/dx=-\sin x

from first principles.

1Start with

\lim_{h\to0}[\cos(x+h)-\cos x]/h

2Expand

\cos(x+h)=\cos x\cos h-\sin x\sin h

3Group to obtain

\cos x\,[(\cos h-1)/h]-\sin x\,[\sin h/h]

4As h tends to zero, the limits of the brackets are 0 and 1, respectively. The result is

0-\sin x

Answer

d(\cos x)/dx=-\sin x

, provided x is measured in radians.

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 $u$ , $v$ , $u^{\prime}$ , $v^{\prime}$ explicitly for product and quotient rules.
Work in radians whenever trigonometric functions are differentiated.

Common misconceptions

Differentiating

\mathrm{e}^{3x}

3x\mathrm{e}^{3x-1}

Exponentials are not powers of

x

\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{3x} = 3\mathrm{e}^{3x}

Differentiating

(2x + 1)^{4}

4(2x + 1)^{3}

Multiply by the inner derivative:

8(2x + 1)^{3}

Writing

\dfrac{\mathrm{d}}{\mathrm{d}x}\cos x = \sin x

\dfrac{\mathrm{d}}{\mathrm{d}x}\cos x = -\sin x

Differentiating a product as the product of the derivatives.

Use

uv^{\prime} + vu^{\prime}

Differentiation rules

\tfrac{\mathrm{d}}{\mathrm{d}x}x^{n} = nx^{n-1}

Power rule.

\tfrac{\mathrm{d}y}{\mathrm{d}x} = \tfrac{\mathrm{d}y}{\mathrm{d}u}\tfrac{\mathrm{d}u}{\mathrm{d}x}

Chain rule.

(uv)^{\prime} = uv^{\prime} + vu^{\prime}

Product rule.

\left(\tfrac{u}{v}\right)^{\prime} = \tfrac{vu^{\prime} - uv^{\prime}}{v^{2}}

Quotient rule.

Stretch yourself

Prove from first principles that the derivative of $\sin x$ is $\cos x$ , given that $\dfrac{\sin h}{h} \to 1$ and $\dfrac{\cos h - 1}{h} \to 0$ 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 $\cos x$ .

Reveal solution

SolutionThe chord gradient is

\dfrac{\sin(x + h) - \sin x}{h} = \dfrac{\sin x\cos h + \cos x\sin h - \sin x}{h}

. Grouping, this is

\sin x\left(\dfrac{\cos h - 1}{h}\right) + \cos x\left(\dfrac{\sin h}{h}\right)

. As

h \to 0

, the first bracket tends to

0

and the second to

1

, so the limit is

\sin x \times 0 + \cos x \times 1 = \cos x

. Hence

\dfrac{\mathrm{d}}{\mathrm{d}x}\sin x = \cos x

. The two given limits are only true in radians — which is precisely why the result requires radians.

\blacksquare

Questions students ask

What does

\dfrac{\mathrm{d}y}{\mathrm{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,

\mathrm{e}

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

Can I avoid the quotient rule?

Yes: write

\dfrac{u}{v}

uv^{-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

\dfrac{\sin h}{h} \to 1

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

\tfrac{\pi}{180}

Key takeaways

The derivative is the limit of the chord gradient as $h \to 0$ .
Power rule $nx^{n-1}$ ; $\mathrm{e}^{kx} \to k\mathrm{e}^{kx}$ ; $\ln x \to \tfrac1x$ ; trig derivatives in radians.
Chain rule for composites: outer derivative times inner derivative.
Product rule $uv^{\prime} + vu^{\prime}$ ; quotient rule $\dfrac{vu^{\prime} - uv^{\prime}}{v^{2}}$ .
Rewrite expressions as powers and factorise results.