| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Inverse function differentiation |
| Difficulty | Standard +0.3 This is a straightforward inverse function differentiation problem requiring the chain rule and the standard result that dx/dy = 1/(dy/dx). Students need to differentiate y = 8tan(2x), then reciprocate and simplify using the identity 1 + tan²(2x) = sec²(2x). While it requires multiple steps, the techniques are standard C3/C4 material with no novel insight needed, making it slightly easier than average. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 16\sec^2(2x)\) | M1 | Achieves \(\frac{dy}{dx} = \lambda\sec^2(2x)\) or implicitly \(1 = \lambda\sec^2(2x)\frac{dx}{dy}\) |
| Inverts: \(\frac{dx}{dy} = \frac{1}{16\sec^2 2x} = \frac{1}{16(1+\tan^2 2x)}\) | dM1 | Two of three processes: (1) reciprocal taken, (2) identity \(1+\tan^2 2x = \sec^2 2x\) attempted, (3) replace \(\tan 2x\) by \(\frac{y}{8}\) |
| \(\frac{dx}{dy} = \frac{A}{B+y^2}\) | ddM1 | All three processes attempted AND fractions eliminated (seen in at least two terms) |
| \(\frac{dx}{dy} = \frac{4}{64+y^2}\) | A1 | cso |
# Question 8:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 16\sec^2(2x)$ | M1 | Achieves $\frac{dy}{dx} = \lambda\sec^2(2x)$ or implicitly $1 = \lambda\sec^2(2x)\frac{dx}{dy}$ |
| Inverts: $\frac{dx}{dy} = \frac{1}{16\sec^2 2x} = \frac{1}{16(1+\tan^2 2x)}$ | dM1 | Two of three processes: (1) reciprocal taken, (2) identity $1+\tan^2 2x = \sec^2 2x$ attempted, (3) replace $\tan 2x$ by $\frac{y}{8}$ |
| $\frac{dx}{dy} = \frac{A}{B+y^2}$ | ddM1 | All three processes attempted AND fractions eliminated (seen in at least two terms) |
| $\frac{dx}{dy} = \frac{4}{64+y^2}$ | A1 | cso |
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\item Given that
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$$y = 8 \tan ( 2 x ) , \quad - \frac { \pi } { 4 } < x < \frac { \pi } { 4 }$$
show that
$$\frac { \mathrm { d } x } { \mathrm {~d} y } = \frac { A } { B + y ^ { 2 } }$$
where $A$ and $B$ are integers to be found.\\
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\hfill \mbox{\textit{Edexcel C34 2018 Q8 [4]}}