A-Level Mathematics May/June 2025 Q5(a): The equation of a curve is xy+y²e^x = 4. Show that dy/dx = (y² - ye^x) / (xe^x + 2y).

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  The equation of the curve is $xy + y^2e^{-x} = 4$.

  We differentiate the equation implicitly with respect to $x$.

  $\frac{d}{dx}(xy + y^2e^{-x}) = \frac{d}{dx}(4)$

  For the term $xy$, we use the product rule: $\frac{d}{dx}(xy) = (1)(y) + (x)(\frac{dy}{dx}) = y + x\frac{dy}{dx}$

  For the term $y^2e^{-x}$, we use the product rule and the chain rule: Let $u = y^2$ and $v = e^{-x}$. $\frac{du}{dx} = 2y\frac{dy}{dx}$ and $\frac{dv}{dx} = -e^{-x}$. $\frac{d}{dx}(y^2e^{-x}) = u\frac{dv}{dx} + v\frac{du}{dx} = (y^2)(-e^{-x}) + (e^{-x})(2y\frac{dy}{dx}) = -y^2e^{-x} + 2ye^{-x}\frac{dy}{dx}$

  The derivative of the constant 4 is 0.

  Combining these results: $(y + x\frac{dy}{dx}) + (-y^2e^{-x} + 2ye^{-x}\frac{dy}{dx}) = 0$

  Now, we rearrange the equation to make $\frac{dy}{dx}$ the subject. $x\frac{dy}{dx} + 2ye^{-x}\frac{dy}{dx} = y^2e^{-x} - y$

  Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(x + 2ye^{-x}) = y^2e^{-x} - y$

  $\frac{dy}{dx} = \frac{y^2e^{-x} - y}{x + 2ye^{-x}}$

  To obtain the required form, we multiply the numerator and the denominator by $e^x$: $\frac{dy}{dx} = \frac{(y^2e^{-x} - y)e^x}{(x + 2ye^{-x})e^x}$

  $\frac{dy}{dx} = \frac{y^2e^{-x}e^x - ye^x}{xe^x + 2ye^{-x}e^x}$

  $\frac{dy}{dx} = \frac{y^2 - ye^x}{xe^x + 2y}$

  This is the required result.

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  How the marks are awarded

  • B1 — Correctly differentiating the term 'xy' using the product rule to get 'y + x dy/dx', as shown in the first line of working.
  • B1 — Correctly differentiating the term 'y²e⁻ˣ' using both the product and chain rules to get '2ye⁻ˣ dy/dx - y²e⁻ˣ', as shown in the second block of differentiation.
  • M1 — Setting the sum of the derivatives of the terms equal to zero and correctly rearranging the equation to collect all 'dy/dx' terms on one side and other terms on the opposite side.
  • A1 — Obtaining the intermediate correct expression for 'dy/dx' and then performing the final, correct algebraic step of multiplying the numerator and denominator by 'eˣ' to arrive at the given answer with no errors.

  Common mistakes

  • Incorrectly differentiating the product 'xy' as just 'x dy/dx'.
  • Errors in differentiating 'y²e⁻ˣ', such as forgetting the minus sign from 'e⁻ˣ' or forgetting to apply the chain rule to 'y²' (i.e., writing '2y' instead of '2y dy/dx').
  • Making sign errors when rearranging the equation to isolate the 'dy/dx' terms, for example writing 'y - y²e⁻ˣ' on the right-hand side.
  • Correctly finding 'dy/dx = (y²e⁻ˣ - y) / (x + 2ye⁻ˣ)' but failing to see the final step of multiplying the numerator and denominator by 'eˣ' to match the given answer.

  Examiner tip: Master implicit differentiation by treating 'y' as a function of 'x' and consistently

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