Edexcel C3 2010 January — Question 4 9 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2010
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeFind derivative of quotient
DifficultyModerate -0.3 Part (i) is a straightforward application of the quotient rule combined with chain rule for ln(x²+1), requiring standard technique with no complications. Part (ii) is a routine implicit differentiation/inverse function problem that follows directly from knowing d/dy(tan y) = sec²y = 1+tan²y. Both parts are standard textbook exercises requiring only direct application of rules with minimal problem-solving.
Spec1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation
4. (i) Given that \(y = \frac { \ln \left( x ^ { 2 } + 1 \right) } { x }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
(ii) Given that \(x = \tan y\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }\).
Question 4:
Part (i):
AnswerMarks Guidance
WorkingMark Guidance
\(y = \frac{\ln(x^2+1)}{x}\)
\(u = \ln(x^2+1) \Rightarrow \frac{du}{dx} = \frac{2x}{x^2+1}\)M1 \(\ln(x^2+1) \to \frac{\text{something}}{x^2+1}\)
A1\(\ln(x^2+1) \to \frac{2x}{x^2+1}\)
Apply quotient rule: \(u=\ln(x^2+1),\ v=x,\ \frac{du}{dx}=\frac{2x}{x^2+1},\ \frac{dv}{dx}=1\)
\(\frac{dy}{dx} = \frac{\left(\frac{2x}{x^2+1}\right)(x) - \ln(x^2+1)}{x^2}\)M1 Applying \(\frac{xu' - \ln(x^2+1)v'}{x^2}\) correctly
A1Correct differentiation with correct bracketing but allow recovery
\(\left\{\frac{dy}{dx} = \frac{2}{(x^2+1)} - \frac{1}{x^2}\ln(x^2+1)\right\}\) {Ignore subsequent working}
Total: (4)
Part (ii):
AnswerMarks Guidance
WorkingMark Guidance
\(x = \tan y\)
\(\frac{dx}{dy} = \sec^2 y\)M1* \(\tan y \to \sec^2 y\) or an attempt to differentiate \(\frac{\sin y}{\cos y}\) using either the quotient rule or product rule
A1\(\frac{dx}{dy} = \sec^2 y\)
\(\frac{dy}{dx} = \frac{1}{\sec^2 y}\ \{=\cos^2 y\}\)dM1* Finding \(\frac{dy}{dx}\) by reciprocating \(\frac{dx}{dy}\)
\(\frac{dy}{dx} = \frac{1}{1+\tan^2 y}\)dM1* For writing down or applying the identity \(\sec^2 y = 1+\tan^2 y\), which must be applied/stated completely in \(y\)
Hence, \(\frac{dy}{dx} = \frac{1}{1+x^2}\) (as required)A1 AG For the correct proof, leading on from the previous line of working
Total: (5)
[9]
## Question 4:

### Part (i):

| Working | Mark | Guidance |
|---------|------|----------|
| $y = \frac{\ln(x^2+1)}{x}$ | | |
| $u = \ln(x^2+1) \Rightarrow \frac{du}{dx} = \frac{2x}{x^2+1}$ | M1 | $\ln(x^2+1) \to \frac{\text{something}}{x^2+1}$ |
| | A1 | $\ln(x^2+1) \to \frac{2x}{x^2+1}$ |
| Apply quotient rule: $u=\ln(x^2+1),\ v=x,\ \frac{du}{dx}=\frac{2x}{x^2+1},\ \frac{dv}{dx}=1$ | | |
| $\frac{dy}{dx} = \frac{\left(\frac{2x}{x^2+1}\right)(x) - \ln(x^2+1)}{x^2}$ | M1 | Applying $\frac{xu' - \ln(x^2+1)v'}{x^2}$ correctly |
| | A1 | Correct differentiation with correct bracketing but allow recovery |
| $\left\{\frac{dy}{dx} = \frac{2}{(x^2+1)} - \frac{1}{x^2}\ln(x^2+1)\right\}$ | | {Ignore subsequent working} |

**Total: (4)**

### Part (ii):

| Working | Mark | Guidance |
|---------|------|----------|
| $x = \tan y$ | | |
| $\frac{dx}{dy} = \sec^2 y$ | M1* | $\tan y \to \sec^2 y$ or an attempt to differentiate $\frac{\sin y}{\cos y}$ using either the quotient rule or product rule |
| | A1 | $\frac{dx}{dy} = \sec^2 y$ |
| $\frac{dy}{dx} = \frac{1}{\sec^2 y}\ \{=\cos^2 y\}$ | dM1* | Finding $\frac{dy}{dx}$ by reciprocating $\frac{dx}{dy}$ |
| $\frac{dy}{dx} = \frac{1}{1+\tan^2 y}$ | dM1* | For writing down or applying the identity $\sec^2 y = 1+\tan^2 y$, which must be applied/stated completely in $y$ |
| Hence, $\frac{dy}{dx} = \frac{1}{1+x^2}$ (as required) | A1 **AG** | For the correct proof, leading on from the previous line of working |

**Total: (5)**

**[9]**
4. (i) Given that $y = \frac { \ln \left( x ^ { 2 } + 1 \right) } { x }$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(ii) Given that $x = \tan y$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }$.\\

\hfill \mbox{\textit{Edexcel C3 2010 Q4 [9]}}
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