Standard +0.3 This is a straightforward implicit differentiation question requiring the product rule and chain rule. Students must differentiate both terms, substitute the given point, and solve for dy/dx. While it involves exponentials and requires careful algebraic manipulation, it follows a standard procedure with no novel insight needed, making it slightly easier than average.
4 A curve has equation \(3 \mathrm { e } ^ { 2 x } y + \mathrm { e } ^ { x } y ^ { 3 } = 14\). Find the gradient of the curve at the point \(( 0,2 )\).
Differentiate \(y^3\) to obtain \(3y^2 \frac{dy}{dx}\)
B1
Use correct product rule at least once
*M1
Obtain \(6e^{2x}y + 3e^{2x} \frac{dy}{dx} + e^x y^3 + 3e^x y^2 \frac{dy}{dx}\) as derivative of LHS
A1
Equate derivative of LHS to zero, substitute \(x = 0\) and \(y = 2\) and find value of \(\frac{dy}{dx}\)
M1(d*M)
Obtain \(-\frac{4}{3}\) or equivalent as final answer
A1
[5]
Differentiate $y^3$ to obtain $3y^2 \frac{dy}{dx}$ | B1 |
Use correct product rule at least once | *M1 |
Obtain $6e^{2x}y + 3e^{2x} \frac{dy}{dx} + e^x y^3 + 3e^x y^2 \frac{dy}{dx}$ as derivative of LHS | A1 |
Equate derivative of LHS to zero, substitute $x = 0$ and $y = 2$ and find value of $\frac{dy}{dx}$ | M1(d*M) |
Obtain $-\frac{4}{3}$ or equivalent as **final answer** | A1 | [5]
4 A curve has equation $3 \mathrm { e } ^ { 2 x } y + \mathrm { e } ^ { x } y ^ { 3 } = 14$. Find the gradient of the curve at the point $( 0,2 )$.
\hfill \mbox{\textit{CAIE P3 2013 Q4 [5]}}