COLLEGE CALCULUS. HELP!
posted by stacy .
1. If f(x)=sin(x^2), Find f'(x) and f'(1)
2. If d/dx(f(3x^3))=9x^2, Calculate f'(x).
3. Let f(x)=cos(6x+6), find f'(x)
4. Let f(x)=2sec(2x), find f'(x)

Looks like an exercise in the chain rule:
f(x) = sin(x^2)
f'(x) = cos(x^2) * 2x = 2x cos(x^2)
f'(1) = 2cos(1)

d/dx(f(3x^3))=9x^2
Let g(x) = f(3x^3)
dg/dx = dg/df * df/dx
9x^2 = dg/df * 9x^2
1 = dg/df
Looks like g(x) = x
so, g(f) = f
dg/dx = df/dx = 9x^2

f(x)=cos(6x+6)
f' = sin(6x+6)*6
= 6sin(6x+6)

f(x)=2sec(2x)
f' = 2sec(2x)tan(2x)*2
= 2sec(2x)tan(2x)