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
 A: 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.
A: $$\begin{align*}
\lim_{h\to0}\frac{f(x+h)-f(x)}h & = \lim_{h\to0}\frac{\sqrt{1+e^{x+h}}-\sqrt{1+e^x}}h \\ 
& = \lim_{h\to0}\frac{\sqrt{1+e^{x+h}}-\sqrt{1+e^x}}h\cdot\dfrac{\sqrt{1+e^{x+h}}+\sqrt{1+e^x}}{\sqrt{1+e^{x+h}}+\sqrt{1+e^x}}\\ 
& = \lim_{h\to0} \dfrac{1+e^{x+h}-1-e^{x}}{h}{\cdot}\dfrac{1}{\sqrt{1+e^{x+h}}+\sqrt{1+e^x}} \\
& = \lim_{h\to0} \dfrac{e^x(e^h-1)}{h}{\cdot}\dfrac{1}{\sqrt{1+e^{x+h}}+\sqrt{1+e^x}} \\
&= e^x\cdot\lim_{h\to0} \dfrac{e^h-1}{h}{\cdot}\dfrac{1}{\sqrt{1+e^{x+h}}+\sqrt{1+e^x}}\\
&= \frac{e^x}{2\sqrt{1+e^{x}}}
\end{align*}$$
For the last part you just have to recognize that you get the derivative of $e^x$ by multiplying by $e^x$: $$e^x\cdot\lim_{h\to0}\frac{e^h-1}h\overset{}=\lim_{h\to0}\dfrac{e^{x+h}-e^h}{h}=e^x$$ and so this means that $\lim\limits_{h\to0}\frac{e^h-1}h=1$. The other result is very obvious: $$\lim_{h\to0}\dfrac{1}{\sqrt{1+e^{x+h}}+\sqrt{1+e^x}}=\dfrac{1}{\sqrt{1+e^{x}}+\sqrt{1+e^x}}=\dfrac{1}{2\sqrt{1+e^{x}}}.$$
A: Remember, the derivative of $\sqrt{x}$ is: $$\frac{1}{2\sqrt{x}}$$
Now, when finding the derivative of $(1+e^x)^{0.5}$, we also need to multiply with the inner derivative, which is $e^x$. Thus we get
$$\frac{dy}{dx} = \frac{1}{(1+e^x)^{0.5}} \cdot e^x$$
Comment if you want some clarification.
A: $$\sqrt{1+e^{x+h}}-\sqrt{1+e^x}=\frac{1+e^{x+h}-(1+e^x)}{\sqrt{1+e^{x+h}}+\sqrt{1+e^x}}$$
$$\implies \frac{\sqrt{1+e^{x+h}}-\sqrt{1+e^x}}h=e^x\cdot\frac1{\sqrt{1+e^{x+h}}+\sqrt{1+e^x}}\cdot\frac{e^h-1}h$$
$$\implies \lim_{h\to0}\frac{\sqrt{1+e^{x+h}}-\sqrt{1+e^x}}h=e^x\cdot\frac1{\lim_{h\to0}(\sqrt{1+e^{x+h}}+\sqrt{1+e^x})}\cdot\lim_{h\to0}\frac{e^h-1}h$$
Can you take it home from here?
A: $$
\begin{align}
\frac{f(x)-f(x_0)}{x-x_0}&=\frac{\sqrt{1+e^x}-\sqrt{1+e^{x_0}}}{x-x_0}=\frac{1+e^x-1-e^{x_0}}{x-x_0}\frac{1}{\sqrt{1+e^x}+\sqrt{1+e^{x_0}}}\\
&= \frac{e^x-e^{x_0}}{x-x_0}\frac{1}{\sqrt{1+e^x}+\sqrt{1+e^{x_0}}} \xrightarrow{x \to x_0}e^{x_0}\frac{1}{2\sqrt{1+e^{x_0}}}
\end{align}
$$
