Question: The definition of derivative of a function $f(x)$ is $$\lim\_{h\to0} \frac{f(x+h)-f(x)}{h}$$

Using this definition, the derivative of $\sin\sqrt{x}$ will be:

$$\lim\_{h\to0} \frac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}$$

$$\lim\_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$

Now i got stuck. How to find the limit or simplify this expression?

I get intuition that we have to use $$\lim\_{x\to0}\frac{\sin x}{x} = 1$$

but that too is leading no where. I am unable to remove h from denominator.

**NOTE**

I know the derivative of $\sin\sqrt{x}$ is $\frac{\cos\sqrt{x}}{2\sqrt{x}}$ using chain rule, but this exercise was given to us for practice using division quotient.

Answer: 
$$\lim\_{h\to 0}\frac{\sin\sqrt{x+h}-\sin\sqrt x}{h}=\lim\_{h\to 0}\frac{\sin\sqrt{x+h}-\sin\sqrt x}{\sqrt{x+h}-\sqrt x}\cdot \frac{\sqrt{x+h}-\sqrt x}{h}$$
$$\underset{u=\sqrt{x+h}}{=}\underbrace{\lim\_{u\to \sqrt x}\frac{\sin(u)-\sin(\sqrt x)}{u-\sqrt x}}\_{=\cos(\sqrt x)}\cdot \underbrace{\lim\_{h\to 0}\frac{\sqrt{x+h}-\sqrt x}{h}}\_{=\frac{1}{2\sqrt x}}=\frac{\cos(\sqrt x)}{2\sqrt x}.$$
You got stuck here

$$2 \cdot \lim\_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$

You can break that one limit up into the product of two limits.

$$\lim\_{h\to 0} \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)
\cdot\lim\_{h\to 0} \frac{\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$

The first limit is easy.

$$\lim\_{h\to 0} \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right) 
 = \cos\left(\frac{\sqrt{x}+\sqrt{x}}{2}\right) = \cos \sqrt x$$

The second isn't that much harder.

\begin{align}
 \lim\_{h\to 0} \frac{\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}
 &= \lim\_{h\to 0} \frac{\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)} 
 {\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)} \cdot
 \frac{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h} \\
 &= \lim\_{h\to 0} \left(1 \cdot
 \dfrac{\sqrt{x+h}-\sqrt{x}}{2h}\right) \\
 &= \lim\_{h\to 0} \dfrac{(x+h)-x}{2h(\sqrt{x+h}+\sqrt{x})} \\
 &= \dfrac{1}{4\sqrt{x}} \\
\end{align}

Putting it all together, you get $\dfrac{\cos\sqrt{x}}{2\sqrt{x}}$