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

Question is asking for gradient at x = 0, y = ln2. e^2y = 5 - e^-x. Differentiation with respect to x: 2e^2y * dy/dx = e^-x . dy/dx = e^-x / 2e^2y. At x = 0, y = ln2 ~ dy/dx = e^0 / 2e^2ln2 = 1 / 2e^ln4 = 1 / 2 * 4 = 1/8. Gradient = 1/8