Question

1.) differentiate: y=sin (3x/2) + cos (3 x/2)



6.) If 3x3 - 3y3 = 48, find y" at point x = 2.

Answer 1

1.
well, to separate it into two problems (which I would not normally do):
let z = 3 x/2 then dz/dx = 3/2 and dx/dz = 2 /3
y = sin z + cos z
dy/dz = cos z - sin z
but
dy/dx = dy/dz dz/dx = 3/2 [ cos z - sin z]
= 3/2 [ cos 3x/2 -sin 3x/2 ]

Answer 2

2. suspect you mean:
If 3x^3 - 3y^3 = 48, find y" at point x = 2.
y^3 = x^3 - 16
y = (x^3 -16)^(1/3)
dy/dx = (1/3) [(x^3-16)^-(2/3)] (3 x^2)

d/dx(dy/dx) = (1/3) { [(x^3-16)^-(2/3)](6x) + (3x^2)(-2/3)(x^3-16)^-(5/3)(3x^2]}
at x = 2
(1/3) { [(8-16)^-(2/3)](12) + (12)(-2/3)(8-16)^-(5/3)(12]} etc