Edexcel Further Maths May 2025 CP2 Q8: Maclaurin Series & Hyperbolic Differentiation
Comprehensive worked solution and examiner mark-scheme analysis by Tutor Sheefa. Focuses on mark allocation, algebraic traps, and step-by-step mathematical reasoning.
The Exam Archetype: Successive Differentiation & Maclaurin Series
Examiners do not invent new mathematics each year — they test consistent structural archetypes. In this paper, Q8 tests your ability to navigate the boundary between conceptual algebra and precise numerical computation.
$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4$$
$$\frac{d}{dx}(\cosh u) = \sinh u \frac{du}{dx}, \quad \frac{d}{dx}(\sinh u) = \cosh u \frac{du}{dx}$$
Examiner Traps & Common Mark-Scheme Penalties
- Product rule slip during third and fourth derivatives.
- Evaluating derivatives at $x=0$: remember $\sinh(0) = 0$ while $\cosh(0) = 1$.
Formal Mathematical Solution
Below is the full step-by-step derivation meeting official mark-scheme criteria for method (M) and accuracy (A) marks:
Successive Derivatives at $x=0$
Odd derivatives vanish due to $\sinh(0) = 0$. Plug even derivatives into the Maclaurin formula.
🎙️ Read Spoken Video Explanation (240 segments, 1204 words) ▾ Expand
Unedited transcript of Tutor Sheefa's spoken audio instructions during the walkthrough:
Let's take a look at question 8. Question 8, we have a differentiation question and also Maclaurin series. So, what we going to do here for question 8, they want us to differentiate Y four times. Notice that we have cos X multiplied by sinh X. That means we have to use product rule. Let's begin. Okay. So, first let's list out. If I have cos X and I want to differentiate, this one will become negative sin X. Next, if I have sin X if I differentiate, I will get cos X. If I have cosh X differentiate, that would be sinh X. Next, if I have sinh X differentiate, that would be cosh X. So, what I'm doing, this is my original and this is after differentiation. So, we will be using this color. When we copy the same expression, [snorts] it will be in purple, but if you differentiate, it will be in blue. So, let's begin from the beginning. So, Y is equal to cos X sinh X. So, dy dx is equal to Let's start by differentiating cos X. We will get negative sin X. And then copy sinh X. Plus copy cos X. Next, we will differentiate sinh X which is cosh x. Simplify this equation, so we will get negative sine x sinh x plus cos x cosh x. We will do the second differentiation now. Okay, let's begin by differentiating negative sine. So, that would be negative cos x copy sinh x copy negative sine x and then differentiate sinh x become cosh x. Next, we will differentiate cos x which is negative sine x copying Oh, wait, because this is differentiation. Okay, so I have to use blue. So, differentiation of cosh that would be negative sine x and then copying cosh x. Okay, and then copying cos x and differentiating cosh x. Let's simplify. So, this would be negative cos x sinh x minus sine x cosh x minus sine x cosh x plus cos x sinh x. These two will eliminate each other. That means our final answer for the second differentiation would be negative two sine x cosh x. Okay, time to do the third differentiation. We'll begin by differentiating negative two sine x. That would be negative two cos x. Copy cosh and then plus copy negative two sine x and differentiate cosh x become shine x. So, our simplification would be negative two cosh x cosh x minus two sine x shine x. Moving on to the fourth differentiation. We'll begin by differentiating negative two cosh. So, that would be positive two sine x. Copying cosh x and then copying negative two cosh x and differentiating cosh x become shine x. Now, we will now differentiate negative two sine x. So, that will become negative two cosh x. Copy shine x plus copy negative two sine x and then differentiate shine x will become cosh x. Let's simplify this equation. So, this will be two sine x cosh x minus two cosh x shine x. -2 cos x sinh x -2 sin x cosh x This and this will eliminate each other. So, we will have -4 cos x sinh x. And what is cos x sinh x? Cos x sinh x is equals to y. So, our answer is the fourth differentiation is equal to -4 y. That's it. We are done with part A. Moving on to part B now. Okay, we want to find the first three non-zero terms for the Maclaurin series. What is Maclaurin series? Okay, we will be using this formula that is given in your formula booklet. So, before we plug in all of this, what we have to do so let's ease our computation. So, what I will do is writing down the value when x is zero for all for trig. So, sin 0 is 0. Cos 0 is 0. Cos 0 is 1. Sinh 0 is e to the power of 0 minus e to the power of 0 over 2, which is equal to 0. And then cosh 0 is equal to e to the power of 0 plus e to the power of 0 over 2, which is 1 + 1 2. 2 over 2 is equal to 1. That's it. Let's start plug this in into our y d y d x until the fourth differentiation. Okay. The first y is equal to cos zero shine zero. Which is equal to one multiplied by zero zero. d y d x is equal to Okay. Negative sign zero shine zero plus cos zero cosh zero. So, what I have is negative zero multiplied by zero plus one multiplied by one, which is equal to one. Moving on to the second differentiation. Second differentiation is equal to negative two sign zero cosh zero. So, this is equal to negative two multiplied by zero multiplied by one, which is equal to zero. The fourth the third differentiation. So, d three y d x cubed equal to negative two cos zero cosh zero minus two sign zero shine zero. So, we will have negative two times one times one minus zero. Which is equal to negative two. Okay. Let me bring this slightly here. Okay. Okay, so the fourth one would be D 4 y D X power of 4 equal to -4 y And what is our y? Our y is zero. So -4 multiplied by zero is equal to zero. They want us to find three non-zero. So we already have one and two. So we need another one. So what we have to do is we have to differentiate again. So let's differentiate this. So D 5 y D X power of 5 is equal to -4. Differentiate y would be D y D X. So -4 multiplied by What is our D y D X? Our D y D X is one. So it would be equals to -4. So now we have three non-zeros value. Let's plug that in into our Maclaurin. Okay. So Maclaurin is equal to y is equal to Okay. 0 + x multiplied by 1 + x [snorts] squared over 2 factorial multiplied by 0 + x cubed over 3 factorial. We're looking at the third one now. Okay. -2 + x to the power of 4 over 4 factorial multiplied by 0 + x to the power of 5 over 5 factorial multiplied by -4. In order to make this Okay, working is more continuous, I'm going to move it down here. So, this would be equal to Y is equal to Okay, here would be zero, here would be X >> [snorts] >> Okay, so 2 over 3 factorial Okay, so that would be 1 over 3 X cubed. That would be another zero. So, now we have Okay, -4 over 5! So, 4 over 5 factorial So, that is -1 over 30 X to the power of 5. With that, we are done with part B as well. Okay, so let's just mark. This is the answer for part B. Okay, and this is the answer for part A. Let me just make this slightly bigger. So, that's it.