Standard +0.3 This is a straightforward implicit differentiation question requiring application of the product rule and chain rule to find dy/dx, then substitution of given coordinates. While it involves exponential functions and requires careful algebraic manipulation, it follows a standard procedure with no novel problem-solving required, making it slightly easier than average.
3 A curve has equation \(6 \mathrm { e } ^ { - x } y ^ { 2 } + \mathrm { e } ^ { 2 x } - 12 y + 7 = 0\).
Find the gradient of the curve at the point \(( \ln 3,2 )\).
Allow correct use of quotient rule. Must have \(\frac{dy}{dx}\) from implicit differentiation.
Obtain \(-6e^{-x}y^2 + 12e^{-x}y\frac{dy}{dx}\)
A1
SOI
Obtain \(+2e^{2x} - 12\frac{dy}{dx}\)
B1
May see separately. \(+2e^{2x} - 12\frac{dy}{dx} + 7\) scores B0.
Rearrange correctly to obtain \(\frac{dy}{dx} = \ldots\)
M1
Correct equation: \(-6e^{-x}y^2 + 12e^{-x}y\frac{dy}{dx} + 2e^{2x} - 12\frac{dy}{dx} = 0\). Must have at least one \(\frac{dy}{dx}\) from implicit differentiation present. May already have used substitution \(\frac{dy}{dx} = \frac{6y^2e^{-x} - 2e^{2x}}{12e^{-x}y - 12}\). M0 for inclusion of an incorrect \(\frac{dy}{dx}\).
Substitute for \(x\) and \(y\) to obtain an exact final answer
M1
Must be at least one \(\frac{dy}{dx}\) from implicit differentiation present.
Obtain \(\frac{5}{2}\) or \(2.5\)
A1
Total
6
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| Use product rule to differentiate $6e^{-x}y^2$ | M1 | Allow correct use of quotient rule. Must have $\frac{dy}{dx}$ from implicit differentiation. |
| Obtain $-6e^{-x}y^2 + 12e^{-x}y\frac{dy}{dx}$ | A1 | SOI |
| Obtain $+2e^{2x} - 12\frac{dy}{dx}$ | B1 | May see separately. $+2e^{2x} - 12\frac{dy}{dx} + 7$ scores B0. |
| Rearrange correctly to obtain $\frac{dy}{dx} = \ldots$ | M1 | Correct equation: $-6e^{-x}y^2 + 12e^{-x}y\frac{dy}{dx} + 2e^{2x} - 12\frac{dy}{dx} = 0$. Must have at least one $\frac{dy}{dx}$ from implicit differentiation present. May already have used substitution $\frac{dy}{dx} = \frac{6y^2e^{-x} - 2e^{2x}}{12e^{-x}y - 12}$. M0 for inclusion of an incorrect $\frac{dy}{dx}$. |
| Substitute for $x$ and $y$ to obtain an exact final answer | M1 | Must be at least one $\frac{dy}{dx}$ from implicit differentiation present. |
| Obtain $\frac{5}{2}$ or $2.5$ | A1 | |
| **Total** | **6** | |
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3 A curve has equation $6 \mathrm { e } ^ { - x } y ^ { 2 } + \mathrm { e } ^ { 2 x } - 12 y + 7 = 0$.\\
Find the gradient of the curve at the point $( \ln 3,2 )$.\\
\hfill \mbox{\textit{CAIE P2 2024 Q3 [6]}}