Question

1. If f(x)=tan3x, find f'(x) and f'(3).
2. let f(x)=3cos(cosx), find f'(x).


i got stuck. i don't know im doing it right way, f'(x)=tan3x*1

Answer 1

f(x) = tan(3x)
use the chain rule to get
f'(x) = sec^2(3x) * 3 = 3sec^2(3x)
f'(3) = 3sec^2(3)

f(x) = 3cos(cos(x))
again, the chain rule works
let f(u) = 3cos(u) and u(x) = cos(x)
df/dx = df/du * du/dx
= -3sin(u) * -sin(x)
= 3sin(x)sin(cos(x))

Answer 2

i don't get the part sec^2<---why use square?