| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Implicit differentiation |
| Difficulty | Moderate -0.3 This is a straightforward two-part question testing standard chain rule and implicit differentiation techniques. Part (i) is routine application of chain rule to a composite function, and part (ii) applies implicit differentiation to the same relationship then verifies equivalence by substitution—both are textbook exercises requiring no problem-solving insight, though the verification step adds minimal challenge beyond pure recall. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{3}(1+3x^2)^{-2/3} \cdot 6x\) | M1 | Chain rule |
| B1 | \(\frac{1}{3}u^{-2/3}\) or \(\frac{1}{3}(1+3x^2)^{-2/3}\) | |
| \(= 2x(1+3x^2)^{-2/3}\) | A1 | o.e. but must be '\(2\)' (not \(6/3\)); mark final answer |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3y^2\frac{dy}{dx} = 6x\) | M1 | \(3y^2\frac{dy}{dx}\) |
| \(dy/dx = 6x/3y^2\) | A1 | \(= 6x\) |
| \(= \frac{2x}{(1+3x^2)^{2/3}} = 2x(1+3x^2)^{-2/3}\) | A1 | |
| E1 | If deriving \(2x(1+3x^2)^{-2/3}\), needs a step of working | |
| [4] |
## Question 8:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{3}(1+3x^2)^{-2/3} \cdot 6x$ | M1 | Chain rule |
| | B1 | $\frac{1}{3}u^{-2/3}$ or $\frac{1}{3}(1+3x^2)^{-2/3}$ |
| $= 2x(1+3x^2)^{-2/3}$ | A1 | o.e. but must be '$2$' (not $6/3$); mark final answer |
| **[3]** | | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3y^2\frac{dy}{dx} = 6x$ | M1 | $3y^2\frac{dy}{dx}$ |
| $dy/dx = 6x/3y^2$ | A1 | $= 6x$ |
| $= \frac{2x}{(1+3x^2)^{2/3}} = 2x(1+3x^2)^{-2/3}$ | A1 | |
| | E1 | If deriving $2x(1+3x^2)^{-2/3}$, needs a step of working |
| **[4]** | | |8 (i) Given that $y = \sqrt [ 3 ] { 1 + 3 x ^ { 2 } }$, use the chain rule to find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$.\\
(ii) Given that $y ^ { 3 } = 1 + 3 x ^ { 2 }$, use implicit differentiation to find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$. Show that this result is equivalent to the result in part (i).
\hfill \mbox{\textit{OCR MEI C3 Q8 [7]}}