Standard +0.3 This is a straightforward implicit differentiation problem requiring product rule and chain rule application, followed by substitution to find a specific point. While it involves multiple algebraic steps, it's a standard C4 exercise with no unusual complications or insight required—slightly easier than average for A-level.
Or \(2x + 2y\frac{dy}{dx} + ky + kx\frac{dy}{dx}\); \(k\) is any positive integer; some terms may appear on RHS with signs reversed
\(k = 2\)
A1
If M0 in middle scheme, SC1 for three terms out of four completely correct with \(k=2\)
\(2y\frac{dy}{dx}\) on RHS from differentiating \(y^2\)
B1
May appear on LHS with sign reversed
\(y^2 + Kxy\frac{dy}{dx}\) on RHS
M1
\(K\) is any positive integer; NB \(K=2\); may appear on LHS with signs reversed
Obtains a value of \(y\) from e.g. \((1+y)^2 = 1\times y^2\) oe
M1
Allow even if follows incorrect manipulation; NB \(y = -0.5\)
Substitution of \(x=1\) and their \(y\) dependent on at least two correct terms seen following differentiation, even if follows subsequent incorrect manipulation
M1
May be implied by \(1+\frac{dy}{dx} = \frac{1}{4} - \frac{dy}{dx}\); or \(\frac{dy}{dx} = \frac{2-1-0.25}{-1-2+1}\); NB \(\frac{dy}{dx} = \frac{2x+2y-y^2}{2xy-2x-2y}\)
\(\frac{dy}{dx} = -\frac{3}{8}\) oe cao
A1 [7]
\(-0.375\)
## Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| LHS is $k(x+y)\!\left(1+\frac{dy}{dx}\right)$ | M1 | Or $2x + 2y\frac{dy}{dx} + ky + kx\frac{dy}{dx}$; $k$ is any positive integer; some terms may appear on RHS with signs reversed |
| $k = 2$ | A1 | If M0 in middle scheme, SC1 for three terms out of four completely correct with $k=2$ |
| $2y\frac{dy}{dx}$ on RHS from differentiating $y^2$ | B1 | May appear on LHS with sign reversed |
| $y^2 + Kxy\frac{dy}{dx}$ on RHS | M1 | $K$ is any positive integer; NB $K=2$; may appear on LHS with signs reversed |
| Obtains a value of $y$ from e.g. $(1+y)^2 = 1\times y^2$ oe | M1 | Allow even if follows incorrect manipulation; NB $y = -0.5$ |
| Substitution of $x=1$ and their $y$ dependent on at least two correct terms seen following differentiation, even if follows subsequent incorrect manipulation | M1 | May be implied by $1+\frac{dy}{dx} = \frac{1}{4} - \frac{dy}{dx}$; or $\frac{dy}{dx} = \frac{2-1-0.25}{-1-2+1}$; NB $\frac{dy}{dx} = \frac{2x+2y-y^2}{2xy-2x-2y}$ |
| $\frac{dy}{dx} = -\frac{3}{8}$ oe cao | A1 **[7]** | $-0.375$ |
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7 A curve has equation $( x + y ) ^ { 2 } = x y ^ { 2 }$. Find the gradient of the curve at the point where $x = 1$.
\hfill \mbox{\textit{OCR C4 2015 Q7 [7]}}