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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?

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You're a brave guy. – Git Gud Jul 24 '13 at 0:25
If you want to cheat somewhat, you could explicitly make use of the chain rule in formal limit procedures and multiply and divide by $\sqrt{(x+h)^2+1}-\sqrt{x^2+1}$. If you don't want to do that, I'm at a loss and @GitGud's sentiments would be mine exactly. – Cameron Williams Jul 24 '13 at 0:26
@Ovi In that case: You are a braver man than I. – Cameron Williams Jul 24 '13 at 0:28
Be sure to wear your hair shirt before working on it :) – David Mitra Jul 24 '13 at 0:28
Heh. That's what lemmas are for. As @Alex suggested, it'd be a heck of a lot easier to establish the chain rule, the power rule, and the derivative of $\sin x$ from first principles than do the "hair shirt" derivation. – Rick Decker Jul 24 '13 at 0:35
up vote 7 down vote accepted

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.

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I love the wording "worth not doing" instead of "not worth doing." – Jonas Meyer Jul 24 '13 at 1:11

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