Differentiation on Edexcel IAL P1 is short, and the syllabus is tight: differentiate $x^n$ and sums of such terms, then use the derivative for gradients, tangents, normals and the second derivative. There is no calculator, so the marks are won or lost on index manipulation and arithmetic with fractions. This post takes the four question types in turn and shows where each mark comes from.
How the marks work
Edexcel mark schemes use three kinds of mark:
- M marks for a correct method, such as reducing a power by one, even if an arithmetic slip follows.
- A marks for accurate answers. They depend on the M mark before them.
- B marks for a correct result that is independent of method.
Two consequences matter in P1. An A mark is lost with its M mark, so a method the examiner cannot see costs both. And a follow-through mark can rescue a later part, but only if the working for the earlier part is visible.
Type 1: rewrite, then differentiate
Most "differentiate" questions test whether you can turn roots and fractions into powers first.
Example 1 · 4 marks
Given $y = 4x^3 - \dfrac{6}{x^2} + 5\sqrt{x}$, find $\dfrac{dy}{dx}$.
Solution
Rewrite every term as a power of $x$: $y = 4x^3 - 6x^{-2} + 5x^{\frac{1}{2}}$.
Differentiate term by term:
$$\frac{dy}{dx} = 12x^2 + 12x^{-3} + \frac{5}{2}x^{-\frac{1}{2}}$$
Marks: M1 for $x^n \to x^{n-1}$ on at least one term, then A1 for each correct term (three terms, so M1 A1 A1 A1).
The sign on a negative power
$-6x^{-2}$ differentiates to $(-6)(-2)x^{-3} = +12x^{-3}$. Two negatives make a positive. Write the multiplication out, because a wrong sign here costs the A mark for that term.
A quotient with a single term underneath splits into separate powers. For example,
$$\frac{x^2-3}{2\sqrt{x}} = \frac{1}{2}x^{\frac{3}{2}} - \frac{3}{2}x^{-\frac{1}{2}},$$
which differentiates to $\frac{3}{4}x^{\frac{1}{2}} + \frac{3}{4}x^{-\frac{3}{2}}$. At $x=4$ this is $\frac{3}{4}(2) + \frac{3}{4}\left(\frac{1}{8}\right) = \frac{51}{32}$. On a non-calculator paper, expect $x$-values like $4$ that make the roots exact.
Type 2: tangents and normals
The routine is always the same four steps: find the point, find the gradient, find the gradient you need, write the line.
Example 2 · 7 marks
The curve $C$ has equation $y = x^3 - 4x^2 + 7$. The point $P$ on $C$ has $x$-coordinate $3$.
(a) Find an equation of the tangent to $C$ at $P$. (b) Find an equation of the normal to $C$ at $P$, in the form $ax+by+c=0$ where $a$, $b$ and $c$ are integers.
Solution
At $x=3$: $y = 27 - 36 + 7 = -2$, so $P$ is $(3, -2)$. (B1)
$\dfrac{dy}{dx} = 3x^2 - 8x$ (M1 A1), so at $x=3$ the gradient is $27 - 24 = 3$. (M1)
(a) Tangent: $y + 2 = 3(x - 3)$, so $y = 3x - 11$. (A1)
(b) The normal has gradient $-\frac{1}{3}$. (M1)
$y + 2 = -\frac{1}{3}(x - 3)$, so $3y + 6 = -x + 3$, giving $x + 3y + 3 = 0$. (A1)
Read the required form. "In the form $ax+by+c=0$ where $a$, $b$ and $c$ are integers" means an answer such as $y = -\frac{1}{3}x - 1$ does not earn the final A mark, even though it describes the same line.
Type 3: find an unknown constant from the gradient
Example 3 · 5 marks
The curve $y = x^2 + \dfrac{k}{x}$, where $k$ is a constant, has gradient $0$ at the point where $x=2$.
(a) Find the value of $k$. (b) Find the value of $\dfrac{d^2y}{dx^2}$ at $x=2$.
Solution
(a) $y = x^2 + kx^{-1}$, so $\dfrac{dy}{dx} = 2x - kx^{-2}$. (M1 A1)
At $x=2$: $4 - \dfrac{k}{4} = 0$, so $k=16$. (A1)
(b) $\dfrac{d^2y}{dx^2} = 2 + 2kx^{-3}$. (M1) At $x=2$ with $k=16$: $2 + \dfrac{32}{8} = 6$. (A1)
Treat $k$ exactly like a number when you differentiate: $\frac{k}{x}$ is $kx^{-1}$, and its derivative is $-kx^{-2}$.
Type 4: the second derivative
In P1 the second derivative is usually a short follow-on, as in Example 3(b): differentiate your $\frac{dy}{dx}$ again and substitute. Keep your first derivative in index form so the second differentiation is one step.
A P1 differentiation checklist
- Rewrite roots and fractions as powers before differentiating, and write that line down: it is where the first M mark often is.
- Keep fractional powers as fractions: $x^{\frac{1}{2}}$, not $x^{0.5}$.
- For a tangent or normal, find the $y$-coordinate from the curve's equation, not from the derivative.
- Normal gradient $= -\dfrac{1}{\text{tangent gradient}}$.
- Give the line in the form the question asks for.
Two more fully marked examples are on the IAL P1 differentiation questions page. Every one of these question types is in the IAL P1 unit page question set, with Edexcel-style mark schemes, and the courses page links the other IAL units.
Practise this topic
FAQ
Is a calculator allowed in Edexcel IAL P1?
No. P1 is the non-calculator paper, so fractional and negative powers, surds and fractions all have to be handled by hand.
What differentiation is in IAL P1?
Differentiating $x^n$ and sums of such terms, gradients, tangents and normals, and the second derivative.
What does "hence" mean in a P1 differentiation question?
Use the result you have just found. A method that ignores it, even a correct one, may not earn the marks.
Exam-style P1 differentiation questions, each with an Edexcel-style mark scheme (M, A and B marks).
Practise IAL P1 →