Moderate -0.3 This is a straightforward implicit differentiation question with two routine parts: substituting coordinates to verify a point (trivial algebra) and finding dy/dx at that point using the product rule and chain rule. The calculation is simpler than average because evaluating at (1,1) makes ln(1)=0, eliminating terms and simplifying the arithmetic significantly.
1 A curve has implicit equation \(y ^ { 2 } + 2 x \ln y = x ^ { 2 }\).
Verify that the point \(( 1,1 )\) lies on the curve, and find the gradient of the curve at this point.
# Question 1
**B1** | Clear evidence of verification needed. At least "1 + 0 = 1" must be correct.
**M1** | $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$
**M1** | $\frac{d}{dx}(2x \ln y) = 2\ln y + \frac{2x}{y}\frac{dy}{dx}$
**A1 cao** | Substituting both $x = 1$ and $y = 1$ into their $\frac{dy}{dx}$ or their equation in $x$, $y$ and $\frac{dy}{dx}$. Must be correct.
**M1** | Not from wrong working.
**A1 cao** | $\frac{dy}{dx} = \frac{2 - 2\ln 1}{2y - 2x/y} = \frac{1}{2}$. Condone $\frac{dy}{dx} = \ldots$ unless pursued.
1 A curve has implicit equation $y ^ { 2 } + 2 x \ln y = x ^ { 2 }$.\\
Verify that the point $( 1,1 )$ lies on the curve, and find the gradient of the curve at this point.
\hfill \mbox{\textit{OCR MEI C3 Q1 [6]}}