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

$$\begin{align}
{(1+e^xe^h)^{0.5}-(1+e^x)^{0.5}\over h}&amp;={(1+e^xe^h)^{0.5}-(1+e^x)^{0.5}\over h}\cdot{(1+e^xe^h)^{0.5}+(1+e^x)^{0.5}\over(1+e^xe^h)^{0.5}+(1+e^x)^{0.5}}\\
&amp;={(1+e^xe^h)-(1+e^x)\over h\left((1+e^xe^h)^{0.5}+(1+e^x)^{0.5}\right)}\\
&amp;={e^x\over(1+e^xe^h)^{0.5}+(1+e^x)^{0.5}}\cdot{e^h-1\over h}
\end{align}$$

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.
$$
\begin{align}
\frac{f(x)-f(x_0)}{x-x_0}&amp;=\frac{\sqrt{1+e^x}-\sqrt{1+e^{x_0}}}{x-x_0}=\frac{1+e^x-1-e^{x_0}}{x-x_0}\frac{1}{\sqrt{1+e^x}+\sqrt{1+e^{x_0}}}\\
&amp;= \frac{e^x-e^{x_0}}{x-x_0}\frac{1}{\sqrt{1+e^x}+\sqrt{1+e^{x_0}}} \xrightarrow{x \to x_0}e^{x_0}\frac{1}{2\sqrt{1+e^{x_0}}}
\end{align}
$$