What is the derivate of $A \cdot\cos{ (B\cdot (x + C))} - D$? $$f(x) = A \cdot \cos{(B\cdot(x+C))} - D$$
$$f'(x) = \text{ ?}$$
I would assume that the derivate is something like this:
$$f'(x) = - A \cdot B \cdot \sin{( \dots )}$$
The thing that troubles me is the inside of the cosine function. The $B$ will stay as it is, but what happens to the $C$?
Thanks in advance!
 A: $$f(x) = A\cos (B (x + C)) - D$$
$$f'(x) = (A\cos (B (x + C)) - D)'$$
$$= (A\cos (B (x + C)))' - D'$$
$$= A(\cos (B (x + C)))' - 0$$
$$= A(\cos (B (x + C))'(B(x+C))'$$
$$=A(-\sin(B(x+C))\cdot B$$
$$=-AB\sin(B(x+C))$$
A: Don't do it all at once. If you were asked to compute
$$ 17 \cos(5 (3\pi + \pi))$$
you wouldn't start by trying to write down the answer. You would first compute $3 \pi + \pi = 4 \pi$, and then $5 (4 \pi) = 20 \pi$, and so forth.
It's the same deal with derivatives: you work through them one piece at a time. The derivative rules are often even presented in a way to make it evident how to do this, but you can always use the chain rule when they're not.
For example, the derivative of
$$ \cos u$$
is
$$ (-1) u' \sin u $$
If you have trouble working it out, it helps to actually define variables and make substitutions to make things easier.
I.E. don't try to find the derviative of
$$ A \cdot \cos{(B\cdot(x+C))} - D $$
Instead, define the variables
$$ v = A \cdot \cos{(B\cdot(x+C))} $$
$$ w = D $$
then compute the derivative of
$$ v - w $$
This will, for example, involve computing $v'$. Repeat the same idea to help organize your work when trying to compute the derivative of $v$.
It's possible to work from the inside out as well. Start by defining
$$ u = x + C$$
and find the derivative $u'$. Then, define
$$ v = B u $$
and compute the derivative $v'$. Then define
$$ w = \cos v $$
and compute $w'$, and so forth. This approach may be more intuitive, since you work through the expression in exactly the same way as if everything was a number and you were just trying to compute a decimal value.
