TITLE: Finding the derivative of $\sin \sqrt {x^2+1}$ from the definition?
QUESTION [5 upvotes]: This means finding $\lim_{h \to 0} \large \large \frac{\sin \sqrt {(x+h)^2+1}-\sin \sqrt {x^2+1}}{h}$ . The only way I could think of to do this is to replace $h$ by some function $f(h)$ such that   $[x+f(h)]^2+1=[x+g(h)]^2$ and this would get rid of one of the square roots, but I was not able to find $f(h)$ . 
Inspired by: Differentiate $\sin \sqrt {x^2+1}$ with respect to x?

REPLY [7 votes]: We are interested in the behaviour of $\frac{\sin(f(x+h))-\sin(f(x))}{h}$ as $h\to 0$.
Note that by the sum (or rather difference) to product formula for $\sin A-\sin B$, the top is equal to 
$$\sin\left(\frac{f(x+h)-f(x)}{2}\right)\cos\left(\frac{f(x+h)+f(x)}{2}\right).$$ 
Tbe cos part will cause no trouble. For the sine part, rationalize the numerator as usual.  
When we rationalize and divide by $h$, the sine part looks like
$$\frac{\sin(g(x,h))}{h}$$
where
$$g(x,h)=\frac{2xh+h^2}{2\sqrt{(x+h)^2+1}+2\sqrt{x^2+1}}.$$
The same trick as the one we use to evaluate $\lim_{t\to 0}\frac{\sin(at)}{t}$ now works, just a little messier to type.
Remark: The above is very much worth not doing.