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
Related Maths A Level answers
All answers ▸(5 + 2(2^0.5))(7 - 3(2^0.5))
A curve has equation y=x^2 + (3k - 4)x + 13 and a line has equation y = 2x + k, where k is constant. Show that the x-coordinate of any point of intersection of the line and curve satisfies the equation: x^2 + 3(k - 2)x + 13 - k = 0
