Differentiation from first principlesEdexcel A-Level Maths: Revision notes

Section 1

The derivative as the gradient of a tangent

The gradient of a curve at a point is the gradient of the tangent there. The derivative $f'(x)$ , also written $\frac{\mathrm{d}y}{\mathrm{d}x}$ , is the gradient function: it gives the gradient at a general point $(x,y)$ . It is also a rate of change: $\frac{\mathrm{d}y}{\mathrm{d}x}$ is the rate of change of $y$ with respect to $x$ . For example, if $s$ is displacement and $t$ is time, $\frac{\mathrm{d}s}{\mathrm{d}t}$ is velocity. The gradient of a tangent is the limit of the gradient of a chord as the second point approaches the first.

Key termsgradienttangentderivativerate of change

Common mistake

Treating $\frac{\mathrm{d}y}{\mathrm{d}x}$ as a fraction $\mathrm{d}\times y\div\mathrm{d}\times x$ . It is a single symbol for the gradient function.

Section 2

Differentiation from first principles

Take a point $P(x,f(x))$ and a nearby point $Q(x+h,f(x+h))$ . The gradient of the chord is $\frac{f(x+h)-f(x)}{h}$ . Then: $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.$ You must expand, simplify and cancel $h$ before letting $h\to0$ . Putting $h=0$ straight away gives $\frac{0}{0}$ . Example, $f(x)=x^{2}$ : $\frac{(x+h)^{2}-x^{2}}{h}=\frac{2xh+h^{2}}{h}=2x+h\to2x$ .

Key termsfirst principleschordlimit

Common mistake

Substituting $h=0$ into the numerator before cancelling, which gives $\frac{0}{0}$ .

Exam tip

After expanding, every term in the numerator should have a factor $h$ to cancel.

Section 3

First principles for x² and x³

For $f(x)=x^{2}$ you need $(x+h)^{2}=x^{2}+2xh+h^{2}$ , giving $f'(x)=2x$ . For $f(x)=x^{3}$ you need $(x+h)^{3}=x^{3}+3x^{2}h+3xh^{2}+h^{3}$ . Then $\frac{(x+h)^{3}-x^{3}}{h}=3x^{2}+3xh+h^{2}$ , and as $h\to0$ , $f'(x)=3x^{2}$ . A constant multiple $kf(x)$ has derivative $kf'(x)$ , and sums and differences are differentiated term by term. Both results match the pattern $\frac{\mathrm{d}}{\mathrm{d}x}x^{n}=nx^{n-1}$ for $n=2,3$ .

Common mistake

Using $(x+h)^{3}=x^{3}+h^{3}$ . The expansion has four terms; use the binomial expansion or Pascal's triangle.

Section 4

The second derivative

Differentiating $f'(x)$ again gives the second derivative, $f''(x)$ or $\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}$ . It is the rate of change of the gradient. If $f(x)=x^{3}$ , then $f'(x)=3x^{2}$ and $f''(x)=6x$ . If $f''(x)>0$ the gradient is increasing; if $f''(x)<0$ the gradient is decreasing. In motion, if $s$ is displacement then $v=\frac{\mathrm{d}s}{\mathrm{d}t}$ and acceleration $a=\frac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}$ is the rate of change of velocity.

Key termssecond derivative

Exam tip

In a context, give the units: if $s$ is in metres and $t$ in seconds, $a$ is in m s $^{-2}$ .

Section 5

Sketching the gradient function

To sketch $y=f'(x)$ from a sketch of $y=f(x)$ , read the gradient of the curve:

Where the curve has a turning point the gradient is $0$ , so $f'(x)$ crosses the $x$ -axis.
Where the curve rises, $f'(x)>0$ ; where it falls, $f'(x)<0$ .
The steeper the curve, the larger the magnitude of $f'(x)$ . For example, a cubic with a local maximum then a local minimum has a quadratic gradient function that is positive, then negative, then positive.

Exam tip

Mark the $x$ -values of the turning points first; they become the roots of the gradient function.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Differentiation from first principles

The point $P(3,9)$ lies on the curve $y=x^{2}$ . The point $Q$ on the same curve has $x$ -coordinate $3+h$ , where $h\neq0$ .
Find the equation of the tangent to the curve at $P$ .2 marks
Let $f(x)=x^{3}$ .
Given that $f'(x)=3x^{2}$ , find $f''(x)$ , and hence find the rate of change of the gradient of the curve $y=f(x)$ when $x=2$ .2 marks
A curve has equation $y=x^{2}$ .
Prove, from first principles, that $\frac{\mathrm{d}y}{\mathrm{d}x}=2x$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).