A-Level Mathematics October/November 2024 Q3: The equation of a curve is ln(x+y) = 3x²y. Find the gradient of the curve at the point…

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  The equation of the curve is given by: $ln(x+y) = 3x^2y$

  We need to find the gradient, $\frac{dy}{dx}$, so we differentiate the equation implicitly with respect to $x$.

  Differentiating the left-hand side (LHS) using the chain rule: $\frac{d}{dx}(ln(x+y)) = \frac{1}{x+y} \cdot \frac{d}{dx}(x+y) = \frac{1}{x+y} (1 + \frac{dy}{dx})$

  Differentiating the right-hand side (RHS) using the product rule: $\frac{d}{dx}(3x^2y) = (3x^2)(\frac{dy}{dx}) + (y)(6x) = 3x^2\frac{dy}{dx} + 6xy$

  Now, we set the differentiated LHS equal to the differentiated RHS: $\frac{1 + \frac{dy}{dx}}{x+y} = 6xy + 3x^2\frac{dy}{dx}$

  To find the gradient at the point (1,0), we substitute $x=1$ and $y=0$ into this equation: $\frac{1 + \frac{dy}{dx}}{1+0} = 6(1)(0) + 3(1)^2\frac{dy}{dx}$

  Simplifying the equation: $1 + \frac{dy}{dx} = 0 + 3\frac{dy}{dx}$ $1 + \frac{dy}{dx} = 3\frac{dy}{dx}$

  Now, we solve for $\frac{dy}{dx}$: $1 = 3\frac{dy}{dx} - \frac{dy}{dx}$ $1 = 2\frac{dy}{dx}$ $\frac{dy}{dx} = \frac{1}{2}$

  The gradient of the curve at the point (1,0) is $\frac{1}{2}$.

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  How the marks are awarded

  • B1 — Correctly differentiating the left-hand side, ln(x+y), using the chain rule to obtain the expression (1+dy/dx)/(x+y).
  • B1 — Correctly differentiating the right-hand side, 3x²y, using the product rule to obtain the expression 6xy + 3x²(dy/dx).
  • M1 — Substituting the coordinates x=1 and y=0 into the fully differentiated equation and attempting to rearrange to solve for dy/dx.
  • A1 — Obtaining the final correct answer of dy/dx = 1/2 after correct working.

  Common mistakes

  • Incorrect application of the chain rule on the LHS, for example differentiating ln(x+y) to get 1/(x+y) without multiplying by (1 + dy/dx).
  • Incorrect application of the product rule on the RHS, for example differentiating 3x²y to get just 6xy or just 3x²(dy/dx).
  • Substituting the point (1,0) into the equation before differentiating, which leads to ln(1) = 0 and does not allow for finding the gradient.
  • Making an algebraic error when solving for dy/dx after substitution, for example rearranging '1 + dy/dx = 3dy/dx' to '1 = 4dy/dx'.

  Examiner tip: When differentiating implicitly, remember to apply the chain rule to any function of y and the product rule whenever x and y terms are multiplied together.

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