How to find derivative of $\sin\sqrt{x}$ using difference quotient? 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.
 A: $$\lim_{h\to 0}\frac{\sin\sqrt{x+h}-\sin\sqrt x}{h}=\lim_{h\to 0}\frac{\sin\sqrt{x+h}-\sin\sqrt x}{\sqrt{x+h}-\sqrt x}\cdot \frac{\sqrt{x+h}-\sqrt x}{h}$$
$$\underset{u=\sqrt{x+h}}{=}\underbrace{\lim_{u\to \sqrt x}\frac{\sin(u)-\sin(\sqrt x)}{u-\sqrt x}}_{=\cos(\sqrt x)}\cdot \underbrace{\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt x}{h}}_{=\frac{1}{2\sqrt x}}=\frac{\cos(\sqrt x)}{2\sqrt x}.$$
A: 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 $\dfrac{\cos\sqrt{x}}{2\sqrt{x}}$
A: 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 :
$$\dfrac{\cos\sqrt{x}}{2\sqrt{x}}$$
