OCR C4 2006 January — Question 2 5 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2006
SessionJanuary
Marks5
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TopicImplicit equations and differentiation
TypeShow dy/dx equals given expression
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring application of the chain rule and product rule to find dy/dx, then algebraic rearrangement to isolate the derivative. It's slightly above average difficulty due to the implicit nature and multiple terms, but follows a standard C4 technique with no novel insight required.
Spec1.07s Parametric and implicit differentiation
2 Given that \(\sin y = x y + x ^ { 2 }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}\)B1
\(\frac{d}{dx}(xy) = x\frac{dy}{dx} + y\) s.o.i.B1 [SR: If \(xy\) taken to LHS, accept \(-x\frac{dy}{dx}+y\) as s.o.i.]
\(\cos y \cdot \frac{dy}{dx} = x\frac{dy}{dx} + y + 2x\) AEFB1
\(f(x,y)\frac{dy}{dx} = g(x,y)\)M1 Regrouping provided > one \(\frac{dy}{dx}\) term
\(\frac{y+2x}{\cos y - x}\) or \(-\frac{y+2x}{x-\cos y}\) or \(\frac{-2x-y}{x-\cos y}\)A1 5 ISW. Answer could imply M1 
# Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$ | B1 | |
| $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$ s.o.i. | B1 | [SR: If $xy$ taken to LHS, accept $-x\frac{dy}{dx}+y$ as s.o.i.] |
| $\cos y \cdot \frac{dy}{dx} = x\frac{dy}{dx} + y + 2x$ AEF | B1 | |
| $f(x,y)\frac{dy}{dx} = g(x,y)$ | M1 | Regrouping provided > one $\frac{dy}{dx}$ term |
| $\frac{y+2x}{\cos y - x}$ or $-\frac{y+2x}{x-\cos y}$ or $\frac{-2x-y}{x-\cos y}$ | A1 | **5** ISW. Answer could imply M1 |
2 Given that $\sin y = x y + x ^ { 2 }$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.

\hfill \mbox{\textit{OCR C4 2006 Q2 [5]}}
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