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

Maths 
bobpursley,
take the differential:
3xy^2 dy+ y^3 dx=5/y dy
dy(2xy^25/y)=y^3 dx
dy/dx= you finish it.
Posted by Anique on .
Find the gradient of the curve xy^3 = 5 ln y at the point (0,1).
take the differential:
3xy^2 dy+ y^3 dx=5/y dy
dy(2xy^25/y)=y^3 dx
dy/dx= you finish it.