| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Implicit or inverse differentiation |
| Difficulty | Moderate -0.3 This is a straightforward multi-part differentiation question testing standard techniques: chain rule for (a)(i), product rule for (a)(ii), and implicit differentiation (or quotient rule then reciprocal) for (b). All are routine applications with no conceptual challenges, making it slightly easier than average, though the variety of techniques prevents it from being trivial. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)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 |
|---|---|---|
| (i) \(= \frac{1}{2}(1-\cos x)^{-\frac{1}{2}} \times \sin x = \frac{\sin x}{2\sqrt{1-\cos x}}\) | M1 A2 | |
| (ii) \(= 3x^2 \times \ln x + x^3 \times \frac{1}{x} = x^3(3\ln x + 1)\) | M1 A2 | |
| (b) \(\frac{dx}{dy} = \frac{1x(3-2y) - (y+1)x(x-2)}{(3-2y)^2} = \frac{5}{(3-2y)^2}\) | M1 A2 | |
| \(\frac{dy}{dx} = 1 + \frac{dx}{dy} = \frac{1}{5}(3-2y)^2\) | M1 A1 | (11) |
**(i)** $= \frac{1}{2}(1-\cos x)^{-\frac{1}{2}} \times \sin x = \frac{\sin x}{2\sqrt{1-\cos x}}$ | M1 A2 |
**(ii)** $= 3x^2 \times \ln x + x^3 \times \frac{1}{x} = x^3(3\ln x + 1)$ | M1 A2 |
**(b)** $\frac{dx}{dy} = \frac{1x(3-2y) - (y+1)x(x-2)}{(3-2y)^2} = \frac{5}{(3-2y)^2}$ | M1 A2 |
$\frac{dy}{dx} = 1 + \frac{dx}{dy} = \frac{1}{5}(3-2y)^2$ | M1 A1 | (11)4. (a) Differentiate each of the following with respect to $x$ and simplify your answers.
\begin{enumerate}[label=(\roman*)]
\item $\sqrt { 1 - \cos x }$
\item $x ^ { 3 } \ln x$\\
(b) Given that
$$x = \frac { y + 1 } { 3 - 2 y } ,$$
find and simplify an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $y$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q4 [11]}}