Could someone please HELP me with this question?

Question

Let (x^2)(y^3) = 5 such that x and y are differentiable functions of t. Determine dy/dt/


a) –[2y(dx/dt) / 3x]
b) –[2x(dx/dt) / 3y^2]
c) –[2xy(dx/dt) / 3]
d) -2y/3x

Answer 1

Using the rule for differentiating products and the chain rule, we get

d(x^2y^3)/dt = d(x^2)/dx y^3 dx/dt + x^2 d(y^3)/dy dy/dt = 2x y^3 dx/dt + 3 x^2 y^2 dy/dt = 0, because x^2y^3 is constant.

Therefore, dy/dt = (-2xy^3)/(3x^2y^2) dx/dt = -2y/(3x) dx/dt. So, the right answer is (a). If you want, you can write dy/dt = –[2y(dx/dt) / 3x], exactly the same thing

Answer 2

Use the product rule:
2x(y^3)dx/dt + 3(x^2)(y^2)dy/dt = 0
dy/dt = -2x(y^3)(dx/dt)/[3(x^2)(y^2)]
= -2y(dx/dt)/(3x)

The answer is (a)

Answer 3

a