Find the gradient at the point (0, ln 2) on the curve with equation e^2y = 5 − e^−x . [4]

Figure out what skills are being tested: implicit differentiation and exponentials and logarithms.e^2y = 5 - e^-x2e^2y(dy/dx) = e^-x(dy/dx) = e^-x/ 2e^2yAt (0, ln2) (dy/dx) = e⁰ / 2e^2ln2 e^2ln2 = 4 as 2ln2 = ln(2²) and e^ln(x)= x (dy/dx) = 1 / 8.