The derivative of sin(t-s)? or f(x)=sin(t-s), f'(x)=?

Question

Ok, the book tells me the answer is -cos(t-s), but I don't understand how they figured it. I believe it uses implicit differentiation, but I don't know how.

Thanks for your help.

But in the d/ds (t-s), where does the t go? If you figure it out it should be cos(t' - 1), right?

Answer 1

For the answer to be correct, we must be differentiating with respect to s. It's just a simple application of the chain rule:
d/ds (sin (t-s)) = cos (t-s) . d/ds (t-s)
= -cos(t-s).

More fully, if we let x(s) = t-s, then
d/ds sin (t-s) = d/ds sin x
= [d/dx (sin x)] . [dx/ds]
= [cos x] d/ds (t-s)
= (cos (t-s)) (-1)
= -cos (t-s).

Answer 2

I believe
df/dx = (df/dz) . (dz/dx) etc. depending on the function and function of function.
Your function has two variables 't' and 's', no 'x'. So I am assuming you are trying to find (df/ds).
Let y=t-s.
(df/dy)=cos(y)=cos(t-s)
(dy/ds)={d(t-s)/ds}=-1
Therefore
(df/ds)=(df/dy)(dy/ds)
=cos(t-s)(-1)
= -cos(t-s)

Good luck.