| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of quotient |
| Difficulty | Moderate -0.3 This is a straightforward multi-part differentiation question testing standard rules (product rule, quotient rule, chain rule, and implicit differentiation). All parts are routine applications of techniques with no problem-solving required, making it slightly easier than average but not trivial since it requires correct application of multiple differentiation rules. |
| Spec | 1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(u = x^3\), \(\frac{du}{dx} = 3x^2\) | ||
| \(v = e^{3x}\), \(\frac{dv}{dx} = 3e^{3x}\) | ||
| \(\frac{dy}{dx} = 3x^2e^{3x} + x^3 \cdot 3e^{3x}\) | M1 A1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(u = 2x\), \(\frac{du}{dx} = 2\) | ||
| \(v = \cos x\), \(\frac{dv}{dx} = -\sin x\) | ||
| \(\frac{dy}{dx} = \frac{2\cos x + 2x\sin x}{\cos^2 x}\) | M1 A1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(u = \tan x\), \(\frac{du}{dx} = \sec^2 x\) | ||
| \(y = u^2\), \(\frac{dy}{du} = 2u\) | ||
| \(\frac{dy}{dx} = 2u\sec^2 x\) | M1 | |
| \(\frac{dy}{dx} = 2\tan x \sec^2 x\) | A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(u = y^2\), \(\frac{du}{dy} = 2y\) | ||
| \(x = \cos u\), \(\frac{dx}{du} = -\sin u\) | M1 | |
| \(\frac{dx}{dy} = -2y\sin y^2\) | A1 | |
| \(\frac{dy}{dx} = \frac{-1}{2y\sin y^2}\) | M1 A1 | (4 marks) |
# Question 4:
## Part (i)
| Answer/Working | Marks | Notes |
|---|---|---|
| $u = x^3$, $\frac{du}{dx} = 3x^2$ | | |
| $v = e^{3x}$, $\frac{dv}{dx} = 3e^{3x}$ | | |
| $\frac{dy}{dx} = 3x^2e^{3x} + x^3 \cdot 3e^{3x}$ | M1 A1 A1 | (3 marks) |
## Part (ii)
| Answer/Working | Marks | Notes |
|---|---|---|
| $u = 2x$, $\frac{du}{dx} = 2$ | | |
| $v = \cos x$, $\frac{dv}{dx} = -\sin x$ | | |
| $\frac{dy}{dx} = \frac{2\cos x + 2x\sin x}{\cos^2 x}$ | M1 A1 A1 | (3 marks) |
## Part (iii)
| Answer/Working | Marks | Notes |
|---|---|---|
| $u = \tan x$, $\frac{du}{dx} = \sec^2 x$ | | |
| $y = u^2$, $\frac{dy}{du} = 2u$ | | |
| $\frac{dy}{dx} = 2u\sec^2 x$ | M1 | |
| $\frac{dy}{dx} = 2\tan x \sec^2 x$ | A1 | (2 marks) |
## Part (iv)
| Answer/Working | Marks | Notes |
|---|---|---|
| $u = y^2$, $\frac{du}{dy} = 2y$ | | |
| $x = \cos u$, $\frac{dx}{du} = -\sin u$ | M1 | |
| $\frac{dx}{dy} = -2y\sin y^2$ | A1 | |
| $\frac{dy}{dx} = \frac{-1}{2y\sin y^2}$ | M1 A1 | (4 marks) |
**Total: 12 marks**
---4. Differentiate with respect to $x$\\
(i) $x ^ { 3 } \mathrm { e } ^ { 3 x }$,\\
(ii) $\frac { 2 x } { \cos x }$,\\
(iii) $\tan ^ { 2 } x$.
Given that $x = \cos y ^ { 2 }$,\\
(iv) find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $y$.\\
\hfill \mbox{\textit{Edexcel C3 Q4 [12]}}