Standard +0.3 This is a straightforward implicit differentiation question requiring application of the product rule and chain rule, followed by algebraic rearrangement. The second part involves setting dy/dx = 0 and solving a simple linear equation. While it requires multiple techniques, these are standard C3 procedures with no novel insight needed, making it slightly easier than average.
2 Given that \(y ^ { 3 } = x y - x ^ { 2 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }\).
Hence show that the curve \(y ^ { 3 } = x y - x ^ { 2 }\) has a stationary point when \(x = \frac { 1 } { 8 }\).
2 Given that $y ^ { 3 } = x y - x ^ { 2 }$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }$.\\
Hence show that the curve $y ^ { 3 } = x y - x ^ { 2 }$ has a stationary point when $x = \frac { 1 } { 8 }$.
\hfill \mbox{\textit{OCR MEI C3 Q2 [7]}}