Originally Posted by

**dwat** Just wanted to check if I am differentiating these functions correctly.

$\displaystyle 1. f(x) = tan (2x)$

$\displaystyle f'(x) = sec^2(2x)\cdot 2$

$\displaystyle f'(x) = 2sec^2(2x)$

The (2x) cannot change

I'd multiply this first

$\displaystyle 2. f(x) = (\cos x-1)(\sin x+1)=\sin x \cos x+\cos x-\sin x -1$

$\displaystyle f'(x)=-\sin^2 x +\cos^2 x-\sin x -\cos x$

which is the same as yours, but less work

But your way was ok here

$\displaystyle f'(x) = -sinx(sinx+1)+cosx(cosx-1)$

$\displaystyle f'(x) = -sin^2x-sinx+cos^2x-cosx$

$\displaystyle 3. f(x) = (cosx-sinx)^2$

$\displaystyle f'(x) = 2(cosx-sinx)(-sinx-cosx)$

This is correct but you should multiply out the terms to see if there is a simpler form, which I doubt.

$\displaystyle 4. f(x) = ln(\cos x)+e^{\cos x}$

The first derivative is wrong, the second is correct

$\displaystyle f'(x) = {-\sin x\over \cos x}+e^{\cos x}(-\sin x)$

which can be reduced to $\displaystyle -\tan x-\sin x e^{\cos x}$.

If a function can be simplified even further, can you please show me.