| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Show derivative equals expression - algebraic/trigonometric identity proof |
| Difficulty | Moderate -0.3 Part (a) is trivial differentiation of standard functions. Part (b)(i) requires quotient rule application with basic simplification. Part (b)(ii) uses chain rule with the result from (b)(i), requiring recognition that cosec x = 1/sin x. All techniques are standard C3 material with straightforward algebraic manipulation, making this slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 3e^{3x} + \frac{1}{x}\) | B1 B1 | One mark each term |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{du}{dx} = \frac{\cos x(1+\cos x) - \sin x(-\sin x)}{(1+\cos x)^2}\) | M1 | Quotient rule applied |
| \(= \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2}\) | A1 | Correct numerator expansion |
| \(= \frac{\cos x + 1}{(1+\cos x)^2} = \frac{1}{1+\cos x}\) | A1 | Simplified to required form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{u}\cdot\frac{du}{dx} = \frac{1+\cos x}{\sin x}\cdot\frac{1}{1+\cos x}\) | M1 | Chain rule using part (i) |
| \(= \frac{1}{\sin x} = \text{cosec}\, x\) | A1 | Correct simplification |
# Question 3:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 3e^{3x} + \frac{1}{x}$ | B1 B1 | One mark each term |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{du}{dx} = \frac{\cos x(1+\cos x) - \sin x(-\sin x)}{(1+\cos x)^2}$ | M1 | Quotient rule applied |
| $= \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2}$ | A1 | Correct numerator expansion |
| $= \frac{\cos x + 1}{(1+\cos x)^2} = \frac{1}{1+\cos x}$ | A1 | Simplified to required form |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{1}{u}\cdot\frac{du}{dx} = \frac{1+\cos x}{\sin x}\cdot\frac{1}{1+\cos x}$ | M1 | Chain rule using part (i) |
| $= \frac{1}{\sin x} = \text{cosec}\, x$ | A1 | Correct simplification |
---3
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when
$$y = \mathrm { e } ^ { 3 x } + \ln x$$
\item \begin{enumerate}[label=(\roman*)]
\item Given that $u = \frac { \sin x } { 1 + \cos x }$, show that $\frac { \mathrm { d } u } { \mathrm {~d} x } = \frac { 1 } { 1 + \cos x }$.
\item Hence show that if $y = \ln \left( \frac { \sin x } { 1 + \cos x } \right)$, then $\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosec } x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2013 Q3 [7]}}