Showing that there exists a certain function We have the following functions:
$f(x) = x-\sin(x); \ \ \ \ g(x) = 1-\cos(x)$
I have to prove that there exists a function h such that h(f(x)) = g(x). But I don't exactly how to show that there exists such a function
It looks like I don't need to find the exact function $h$, only to show the existence of one. I already proved that $f(x)$ is bijective. But what about that function $h$ which should exists? What can I do? Can I use the fact that $f$ is bijective?
I also trying to find the derivative of $f(x)$. Yes, I am able to do that with my own knowledge of derivatives (and it is equal to $g(x)$), but I am also trying with the definition of the derivative. I tried the following:
$f(x+h) = x+h - \sin(x+h)$, but then I have write is as $f(x) + Df_x + r(h)$ where $r(h) \to 0 \text{ as } h \to 0$. But how can I rewrite $f(x+h)$ to get in into those terms?
 A: Since $f$ is bijective just take $h(x) = g(f^{-1}(x))$. Then
$$h(f(x)) = g(f^{-1}(f(x))) = g(x) $$
as required.
A: I will try to answer your second query about proving from first principles that
$$\frac{d}{dx}(x-\sin x)=1-\cos x$$
Recall that the derivative, $f'(x)$, of a function $f(x)$, is defined as
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
For $f(x)=x-\sin x$
$$\frac{f(x+h)-f(x)}{h}=\frac{((x+h)-\sin(x+h))-(x-\sin x)}{h}=\frac{h}{h}-\frac{\sin(x+h)-\sin x}{h}$$
$$=1-\frac{\sin(x+h)-\sin x}{h}$$
Hence,
$$f'(x)=\lim_{h\to0}\left(1-\frac{\sin(x+h)-\sin x}{h}\right)=1-\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}$$
Now to focus on $\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}$:
$$\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}=\lim_{h\to0}\frac{\sin x \cos h+\sin h \cos x-\sin x}{h}$$
$$=\lim_{h\to0}\left(\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right)$$
Now we use the facts that
$$\lim_{h\to0}\frac{\cos h-1}{h}=0,~~~\lim_{h\to0}\frac{\sin h}{h}=1$$
Hence we are left with
$$\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}=0+\cos x \times 1=\cos x$$
So finally,
$$f'(x)=x-\cos x$$ as required.

For a proof for the $2$ standard limits used above, see the Wikipedia article on the squeeze theorem (the second example): https://en.wikipedia.org/wiki/Squeeze_theorem
