Question

How do you differentiate `f(x) = cos (3x -3)`?

Answer 1

`-3\sin(3x-3)`

Explanation:

`f(x)=\cos(3x-3)`
`f'(x)=d/dx(\cos(3x-3))`

Applying chain rule,
`d/dxf(g(x))=d/dxg(x)*d/dx(f(g(x))`

Thus...
`f'(x)=d/dx(3x-3)*d/dx(\cos(3x-3))`

Simplified, `f'(x)=3*(-\sin(3x-3))`
`=-3\sin(3x-3)`