The equation of a curve is ye^(2x) + y²e^x = 6. Find the gradient of the curve at the point where y = 1.

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

1. [M1] Substitute y =1 and obtain e^x = a, where a>0. Must come from a quadratic in e^x. (e^(2x) + e^x - 6 = 0). Ignore any negative solution.
2. [A1] Obtain e^x = 2 only. Or equivalent e.g. x = ln 2. Condone x ≈ 0.693.....
3. [B1] State or imply d/dx(ye^(2x)) = 2ye^(2x) + e^(2x)dy/dx. Accept y' for dy/dx.
4. [B1] State or imply d/dx(y²e^x) = y²e^x + 2ye^x dy/dx. Accept y' for dy/dx.
5. [M1] Differentiate RHS of given equation to obtain zero (could be implied by subsequent work), substitute for x and y and obtain dy/dx = ... Independent. [2 * 4 + 4 + 2 + 4 * (dy/dx) = 0]. NB: Could also rearrange to obtain dy/dx then substitute. Both steps are needed for M1.
6. [A1] Obtain -5/4 or -1.25. Correct answer from correct working only. Accept -10/8. -1.2499 is A0.