Standard +0.3 This is a straightforward implicit differentiation question requiring the product rule and chain rule, with a clear substitution point. While it involves exponential functions, it's a standard textbook exercise with no novel insight required—slightly easier than average due to being a single-step problem once differentiated.
4 The equation of a curve is \(\mathrm { ye } ^ { 2 \mathrm { x } } + \mathrm { y } ^ { 2 } \mathrm { e } ^ { \mathrm { x } } = 6\).
Find the gradient of the curve at the point where \(y = 1\).
Substitute \(y = 1\) and obtain \(e^x = a\), where \(a > 0\)
M1
Must come from a quadratic in \(e^x\). \(\left(e^{2x} + e^x - 6 = 0\right)\). Ignore any negative solution
Obtain \(e^x = 2\) only
A1
Or equivalent e.g. \(x = \ln 2\). Condone \(x = 0.693\ldots\)
State or imply \(\frac{d}{dx}\left(ye^{2x}\right) = 2ye^{2x} + e^{2x}\frac{dy}{dx}\)
B1
Accept \(y'\) for \(\frac{dy}{dx}\)
State or imply \(\frac{d}{dx}\left(y^2 e^x\right) = y^2 e^x + 2ye^x\frac{dy}{dx}\)
B1
Accept \(y'\) for \(\frac{dy}{dx}\)
Differentiate RHS of given equation to obtain zero (could be implied by subsequent work), substitute for \(x\) and \(y\) and obtain \(\frac{dy}{dx} = \ldots\)
M1
Independent. \(\left[2\times4 + 4\frac{dy}{dx} + 2 + 4\frac{dy}{dx} = 0\right]\). NB: Could also rearrange to obtain \(\frac{dy}{dx}\) then substitute. Both steps are needed for M1
Obtain \(-\frac{5}{4}\) or \(-1.25\)
A1
Correct answer from correct working only. Accept \(\frac{-10}{8}\). \(-1.2499\) is A0
Alternative Method for Question 4 (Dividing through by \(e^x\)):
Answer
Marks
Guidance
Answer
Mark
Guidance
Substitute \(y = 1\) and obtain \(e^x = a\), where \(a > 0\)
(M1)
Must come from a quadratic in \(e^x\). \(\left(e^{2x} + e^x - 6 = 0\right)\). Ignore any negative solution
Obtain \(e^x = 2\) only
(A1)
Or equivalent e.g. \(x = \ln 2\). Condone \(x = 0.693\ldots\)
State or imply \(\frac{d}{dx}\left(y^2\right) = 2y\frac{dy}{dx}\)
(B1)
Accept \(y'\) for \(\frac{dy}{dx}\)
State or imply \(\frac{d}{dx}\left(ye^x\right) = ye^x + e^x\frac{dy}{dx}\)
(B1)
Accept \(y'\) for \(\frac{dy}{dx}\)
Differentiate RHS of given equation to obtain \(-6e^{-x}\), substitute for \(x\) and \(y\) and obtain \(\frac{dy}{dx} = \ldots\)
(M1)
Independent. \(\left[1\times2 + 2\frac{dy}{dx} + 2\frac{dy}{dx} = -\frac{6}{2}\right]\). NB: Could also rearrange to obtain \(\frac{dy}{dx}\) then substitute. Both steps are needed for M1
Obtain \(-\frac{5}{4}\) or \(-1.25\)
(A1)
Correct answer from correct working only. Accept \(\frac{-10}{8}\). \(-1.2499\) is A0
## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $y = 1$ and obtain $e^x = a$, where $a > 0$ | M1 | Must come from a quadratic in $e^x$. $\left(e^{2x} + e^x - 6 = 0\right)$. Ignore any negative solution |
| Obtain $e^x = 2$ only | A1 | Or equivalent e.g. $x = \ln 2$. Condone $x = 0.693\ldots$ |
| State or imply $\frac{d}{dx}\left(ye^{2x}\right) = 2ye^{2x} + e^{2x}\frac{dy}{dx}$ | B1 | Accept $y'$ for $\frac{dy}{dx}$ |
| State or imply $\frac{d}{dx}\left(y^2 e^x\right) = y^2 e^x + 2ye^x\frac{dy}{dx}$ | B1 | Accept $y'$ for $\frac{dy}{dx}$ |
| Differentiate RHS of given equation to obtain zero (could be implied by subsequent work), substitute for $x$ and $y$ and obtain $\frac{dy}{dx} = \ldots$ | M1 | Independent. $\left[2\times4 + 4\frac{dy}{dx} + 2 + 4\frac{dy}{dx} = 0\right]$. NB: Could also rearrange to obtain $\frac{dy}{dx}$ then substitute. Both steps are needed for M1 |
| Obtain $-\frac{5}{4}$ or $-1.25$ | A1 | Correct answer from correct working only. Accept $\frac{-10}{8}$. $-1.2499$ is A0 |
**Alternative Method for Question 4 (Dividing through by $e^x$):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $y = 1$ and obtain $e^x = a$, where $a > 0$ | (M1) | Must come from a quadratic in $e^x$. $\left(e^{2x} + e^x - 6 = 0\right)$. Ignore any negative solution |
| Obtain $e^x = 2$ only | (A1) | Or equivalent e.g. $x = \ln 2$. Condone $x = 0.693\ldots$ |
| State or imply $\frac{d}{dx}\left(y^2\right) = 2y\frac{dy}{dx}$ | (B1) | Accept $y'$ for $\frac{dy}{dx}$ |
| State or imply $\frac{d}{dx}\left(ye^x\right) = ye^x + e^x\frac{dy}{dx}$ | (B1) | Accept $y'$ for $\frac{dy}{dx}$ |
| Differentiate RHS of given equation to obtain $-6e^{-x}$, substitute for $x$ and $y$ and obtain $\frac{dy}{dx} = \ldots$ | (M1) | Independent. $\left[1\times2 + 2\frac{dy}{dx} + 2\frac{dy}{dx} = -\frac{6}{2}\right]$. NB: Could also rearrange to obtain $\frac{dy}{dx}$ then substitute. Both steps are needed for M1 |
| Obtain $-\frac{5}{4}$ or $-1.25$ | (A1) | Correct answer from correct working only. Accept $\frac{-10}{8}$. $-1.2499$ is A0 |
4 The equation of a curve is $\mathrm { ye } ^ { 2 \mathrm { x } } + \mathrm { y } ^ { 2 } \mathrm { e } ^ { \mathrm { x } } = 6$.\\
Find the gradient of the curve at the point where $y = 1$.\\
\hfill \mbox{\textit{CAIE P3 2024 Q4 [6]}}