Moderate -0.8 This is a straightforward application of the quotient rule to differentiate an exponential function, followed by direct substitution of a given x-value. The algebra is routine (substituting ln 3 simplifies nicely to give 9/50), and the question explicitly tells students what answer to reach, removing any uncertainty about whether their work is correct.
2 The equation of a curve is \(y = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }\). Show that the gradient of the curve at the point for which \(x = \ln 3\) is \(\frac { 9 } { 50 }\).
Use correct quotient or product rule or equivalent
M1
Obtain \(\frac{(1+e^{2x}).2e^{2x} - e^{2x}.2e^{2x}}{(1+e^{2x})^2}\) or equivalent
A1
Substitute \(x = \ln 3\) into attempt at first derivative and show use of relevant logarithm property at least once in a correct context
M1
Confirm given answer \(\frac{9}{50}\) legitimately
A1
[4]
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use correct quotient or product rule or equivalent | M1 | |
| Obtain $\frac{(1+e^{2x}).2e^{2x} - e^{2x}.2e^{2x}}{(1+e^{2x})^2}$ or equivalent | A1 | |
| Substitute $x = \ln 3$ into attempt at first derivative and show use of relevant logarithm property at least once in a correct context | M1 | |
| Confirm given answer $\frac{9}{50}$ legitimately | A1 | [4] |
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2 The equation of a curve is $y = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }$. Show that the gradient of the curve at the point for which $x = \ln 3$ is $\frac { 9 } { 50 }$.
\hfill \mbox{\textit{CAIE P3 2011 Q2 [4]}}