Derivative of $\; y={(1+e^x)}^{0.5}\; $ using the definition of the derivative $$y={(1+e^x)}^{0.5} =f(x)$$
$$\frac{dy}{dx}= \lim_{h\to0}\frac {f(x+h)-f(x)}{h}$$
My attempt I got down to 
$$\lim_{h\to0}\frac{(1+e^xe^h)^{0.5}-(1+e^x)^{0.5}}{h}$$
I can't see where to go from here

Provided you know
$$\lim_{h\to0}{e^h-1\over h}=1$$
you can take what you got to and write
$$\begin{align}
{(1+e^xe^h)^{0.5}-(1+e^x)^{0.5}\over h}&={(1+e^xe^h)^{0.5}-(1+e^x)^{0.5}\over h}\cdot{(1+e^xe^h)^{0.5}+(1+e^x)^{0.5}\over(1+e^xe^h)^{0.5}+(1+e^x)^{0.5}}\\
&={(1+e^xe^h)-(1+e^x)\over h\left((1+e^xe^h)^{0.5}+(1+e^x)^{0.5}\right)}\\
&={e^x\over(1+e^xe^h)^{0.5}+(1+e^x)^{0.5}}\cdot{e^h-1\over h}
\end{align}$$
When you take the limit now, you get $e^x/2(1+e^x)^{0.5}$.
The multiplication step is a standard trick for working with a square root.  Computing the limit of $(e^h-1)/h$ requires more than the definition of the derivative; it requires some definition of the exponential function.
