Post a New Question


posted by on .

Find the gradient of the curve xy^3 = 5 ln y at the point (0,1).

  • Maths - ,

    take the differential:

    3xy^2 dy+ y^3 dx=5/y dy

    dy(2xy^2-5/y)=-y^3 dx

    dy/dx= you finish it.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question