Using the Definition of Derivative to Derive the Derivative of a Function Assume $f$ is differentiable at point $a$, and $f(a)&gt;0$. Determine the derivative of 
$$g(x)=x\sqrt{f(x)}$$
in terms of $f'(a)$.

What I have done is substituting whatever is given and I don't really know how to continue already.

Really appreciate your help!
Okay using chain rule you would have:
$$\frac{d}{dx}g(x)=\frac{d}{dx}(x\cdot \sqrt{f(x)})$$
$$=1\cdot\sqrt{f(x)}+x \cdot \frac{f'(x)}{2\sqrt{f(x)}} $$
Using the actual definition of the derivative via limits, this is what we do.
$$\lim_{h\rightarrow 0} \frac{g(x+h)-g(x)}{(x+h)-(x)}=$$
$$\lim_{h\rightarrow 0} \frac{(x+h)\cdot \sqrt{f(x+h)}-x\cdot \sqrt{f(x)}}{(x+h)-(x)}=$$
$$\lim_{h\rightarrow 0} \frac{(x+h)\cdot \sqrt{f(x+h)}-x\cdot \sqrt{f(x)}}{(x+h)-(x)}=$$
Now we expand numerator to "split" the interval of the limit.
$$\lim_{h\rightarrow 0} \frac{(x+h)\cdot \sqrt{f(x+h)}-(x+\frac{h}{2})\sqrt{x+\frac{h}{2}}+(x+\frac{h}{2})\sqrt{x+\frac{h}{2}}-x\cdot \sqrt{f(x)}}{(x+h)-(x)}=$$
going to have to finish this later.
Okay using chain rule you would have:
$$\frac{d}{dx}g(x)=\frac{d}{dx}(x\cdot \sqrt{f(x)})$$
$$=1\cdot\sqrt{f(x)}+x \cdot \frac{f'(x)}{2\sqrt{f(x)}} $$
Using the actual definition of the derivative via limits, this is what we do.
$$\lim_{h\rightarrow 0} \frac{g(x+h)-g(x)}{(x+h)-(x)}=$$
$$\lim_{h\rightarrow 0} \frac{(x+h)\cdot \sqrt{f(x+h)}-x\cdot \sqrt{f(x)}}{(x+h)-(x)}=$$
$$\lim_{h\rightarrow 0} \frac{(x+h)\cdot \sqrt{f(x+h)}-x\cdot \sqrt{f(x)}}{(x+h)-(x)}=$$
Now we expand numerator to "split" the interval of the limit.
$$\lim_{h\rightarrow 0} \frac{(x+h)\cdot \sqrt{f(x+h)}-(x+\frac{h}{2})\sqrt{x+\frac{h}{2}}+(x+\frac{h}{2})\sqrt{x+\frac{h}{2}}-x\cdot \sqrt{f(x)}}{(x+h)-(x)}=$$
going to have to finish this later.