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 are right. First multiply and divide by the expression in $\sin()$
$$\lim\_{h\rightarrow 0} \dfrac{2\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}\times\dfrac{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}$$

Now, as $h \rightarrow 0, \ \ \ \left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)\rightarrow 0$

So we can use $\ \ \lim\_{\lambda \rightarrow 0} \dfrac{\sin\lambda}{\lambda} = 1$

So it reduces to:
$$\lim\_{h\rightarrow 0} \dfrac{\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)}{h}\times\left(\sqrt{x+h}-\sqrt{x}\right)$$

Multiply and divide by $\left(\sqrt{x+h}+\sqrt{x}\right)$

$$\lim\_{h\rightarrow 0} \dfrac{\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)}{h}\times\dfrac{\left((\sqrt{x+h})^2-(\sqrt{x})^2\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)}$$

$$ = \lim\_{h\rightarrow 0} \dfrac{\cos\left(\dfrac{\sqrt{x+h}+\sqrt{x}}{2}\right)}{h}\times\dfrac{h}{\left(\sqrt{x+h}+\sqrt{x}\right)}$$

Apply the limit and you get :

$$\dfrac{\cos\sqrt{x}}{2\sqrt{x}}$$