Standard +0.3 This is a straightforward implicit differentiation question requiring product rule and chain rule application, followed by substitution of given coordinates. While it involves multiple terms and algebraic manipulation, it's a standard textbook exercise with no novel problem-solving required, making it slightly easier than average.
State or imply \(3y^2 + 6xy\frac{dy}{dx}\) as derivative of \(3xy^2\)
B1
State or imply \(3y^2\frac{dy}{dx}\) as derivative of \(y^3\)
B1
Equate derivative of LHS to zero, substitute \((1, 3)\) and find the gradient
M1
\(\left(\frac{dy}{dx} = \frac{x^2+y^2}{y^2-2xy}\right)\) For incorrect derivative need to see the substitution
Obtain final answer \(\frac{10}{3}\) or equivalent
A1
3.33 or better. Allow \(\frac{30}{9}\) ISW after correct answer seen
Total
4
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $3y^2 + 6xy\frac{dy}{dx}$ as derivative of $3xy^2$ | B1 | |
| State or imply $3y^2\frac{dy}{dx}$ as derivative of $y^3$ | B1 | |
| Equate derivative of LHS to zero, substitute $(1, 3)$ and find the gradient | M1 | $\left(\frac{dy}{dx} = \frac{x^2+y^2}{y^2-2xy}\right)$ For incorrect derivative need to see the substitution |
| Obtain final answer $\frac{10}{3}$ or equivalent | A1 | 3.33 or better. Allow $\frac{30}{9}$ ISW after correct answer seen |
| **Total** | **4** | |
3 Find the gradient of the curve $x ^ { 3 } + 3 x y ^ { 2 } - y ^ { 3 } = 1$ at the point with coordinates $( 1,3 )$.\\
\hfill \mbox{\textit{CAIE P3 2019 Q3 [4]}}