 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/4/a/solution

3 y^{2} y^{'} + 2 x = 0

. At

x = 1

, the real curve has

y^{3} = 1

, hence

y = 1

. Therefore

\frac{d y}{d x}_{(1, 1)} = - \frac{2}{3} .

 New to topics? Read the docs here!