Gradient of a function?

Question

Fi nd the gradient of the function

(3 t^3/2)/(2 sin2t)

Give the answer as a product of factors

Answer 1

y = (3t^3/2) / [2sin(2t)] = 3t^3 / [4sin(2t)]

Letting u = 3t^3 and v = 4sin(2t) and using

d(u/v)/dt = [v(du/dt) - u(dv/dt)] / v^2, gives :

dy/dt = {3t^2 / [4sin^2(2t)]} [3sin(2t) - 2tcos(2t)]

being a product of factors.

Answer 2

y = (3t^(3/2))/(2sin2t)

to find the gradient function you must differentiate the function. To do this you will need to seperate the function into the divsion of two simpler functions where u = (3 t^3/2) and v = 2sin2t

to differentiate y you now need to differentiate u/v and to do this you can use the quotient rule:

d(u/v)/dt = (v du/dx - u dv/dx) / v^2

Tod this you must first work out the derivatives of u and v. For u it is fairly easy and for v simply use the rule:

d(a sin bt)/dt = ab cos bt (where a and b are constants)