Unlimited generated questions on this topic, at up to four difficulty levels, with full mark schemes. Notes and worksheets are open to everyone. Practice needs a free account.
Or take a quick quiz: 7 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
What a derivative is: the gradient of the tangent, found as the limit of the gradient of a chord. You need to differentiate simple powers of \(x\) from first principles, sketch gradient functions and use the second derivative.
The chord from \((x, f(x))\) to \((x + h, f(x + h))\) has gradient \(\dfrac{f(x + h) - f(x)}{h}\)
As \(h \to 0\) the chord becomes the tangent, so \(f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}\)
As \(h \to 0\) the chord (red) turns into the tangent (blue), so its gradient tends to \(f'(x)\)
\(f'(x)\), or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), is the gradient of the tangent at a general point, and the rate of change of \(y\) with respect to \(x\)
First principles method: write the limit, expand \(f(x + h)\), cancel, divide every term by \(h\), and only then let \(h \to 0\)
Sketching the gradient function
Where \(y = f(x)\) has a stationary point, the graph of \(y = f'(x)\) meets the x-axis.
Where \(f(x)\) is increasing, \(f'(x)\) is above the x-axis; where it is decreasing, below.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Prove from first principles (powers of \(x\))
4 to 5 marksGrade B
Write the gradient of the chord: \(\dfrac{f(x + h) - f(x)}{h}\)
Expand \(f(x + h)\) and cancel the terms without \(h\)
Divide every remaining term by \(h\)
Let \(h \to 0\), using limit notation, and state the result.
Example. Prove, from first principles, that the derivative of \(2x^3\) is \(6x^2\)
As \(h \to 0\),\(6xh + 2h^2 \to 0\), so the derivative of \(2x^3\) is \(6x^2\)
A1
Don’t lose marks
Putting \(h = 0\) before dividing by \(h\), which gives \(\dfrac{0}{0}\)
Dropping the 'lim' notation, or writing '\(h = 0\)' instead of '\(h \to 0\)': the final mark is often lost.
Using the rule \(nx^{n-1}\) in a first principles question: this scores no marks.
More tips
Memory tricks
First principles layout: limit, expand, cancel, divide by \(h\), let \(h \to 0\)
Differentiating lowers the degree by one: a cubic curve has a quadratic gradient function, so its gradient graph is a parabola.
Exam technique
Write \(\lim_{h \to 0}\) on every line until you actually take the limit, and finish with a concluding sentence.
Give intervals as inequalities, or in set notation if the question asks, e.g. \(\{x : x \leq 2\}\)
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Interpret a derivative as gradient and rate\(f'(x)\) is the gradient of the tangent at \((x, f(x))\) and the rate of change of \(y\) with respect to \(x\)
Grade D
Find a second derivativeDifferentiate \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) again to get \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\), the rate of change of the gradient.
Grade C
Sketch the gradient function of a curveWhere the curve has a stationary point, \(y = f'(x)\) meets the x-axis; where the curve increases, \(f'(x) > 0\)
Grade B
Differentiate \(x^2\) or \(x^3\) from first principlesExpand \(f(x + h)\), simplify \(\dfrac{f(x+h) - f(x)}{h}\), then let \(h \to 0\)
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
Given that \(f(x) = x^{2} + 9x - 2\), prove, from first principles, that \(f'(x) = 2x + 9\).[4]
Show the answer and mark scheme
Answer: Proof, e.g. \(\frac{f(x + h) - f(x)}{h} = 2x + 9 + h \to 2x + 9\) as \(h \to 0\).
M1 for forming the gradient of a chord, \(\frac{f(x + h) - f(x)}{h} = \frac{(x + h)^{2} + 9(x + h) - 2 - (x^{2} + 9x - 2)}{h}\)
A1 for a correct numerator, \(2xh + h^{2} + 9h\)
A1 for dividing by \(h\) to obtain \(2x + 9 + h\)
A1* for a complete proof: as \(h \to 0\), \(2x + 9 + h \to 2x + 9\), so the derivative is \(2x + 9\) (cso)
(a) Prove, from first principles, that \(\frac{\mathrm{d}y}{\mathrm{d}x} = 12x^{3} + 8\).[5]
(b) Using your answer to part (a), find the equation of the tangent to \(C\) at the point where \(x = -1\). Give your answer in the form \(y = mx + c\).[3]