Differentiate $\sqrt{1+e^x}$ using the definition of a derivative This is the progress I've made so far.
$$\lim_{h \to 0} \frac{\sqrt{1+e^{x+h}}-\sqrt{1+e^{x}}}{h}$$
$$= \lim_{h \to 0} \frac{\left(1+e^{x+h}\right)-\left(1+e^{x}\right)}{h\left(\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}\right)}$$
$$= \lim_{h \to 0} \frac{e^x\left(e^h-1\right)}{h\left(\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}\right)}$$
$$= \lim_{h \to 0} \frac{e^x\left(e^h-1\right)}{h\left(\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}\right)}$$
$$= \lim_{h \to 0} \frac{e^x}{\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}} \frac{e^h-1}{h}$$
I can see how as h tends to 0 the fraction on the left will tend to the desired result, but I'm not sure how to deal with the fraction on the right.
 A: hint use $\lim_{h\to 0}(1+h)^{1/h}=e$ so we can use this approximation and get $e^h \approx (1+h)$ and then substitute value of $h=0$ and get the answer as $1$!
A: In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities
$$1+x\le e^x\le \frac{1}{1-x} \tag 1$$
for $x<1$.  Then, we have
$$1\le \frac{e^h-1}{h}\le \frac{1}{1-h}$$
whereupon applying the squeeze theorem, we obtain the coveted limit.
And in THIS ANSWER, I showed using the inequalities in $(1)$ that 
$$\lim_{h\to 0}\frac{b^h-1}{h}=\log(b)$$
for $b>0$.  Letting $b=e$ immediately gives 
$$\lim_{h\to 0}\frac{e^h-1}{h}=1$$
A: $$\lim_{h \to 0} \frac{e^x}{\left(\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}\right)} \cdot \frac{\left(e^h-1\right)}{h}$$
Since $$e^h=\frac{h^0}{0!}+\frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+......$$
$$e^h=1+\frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+......$$
$$e^h-1= \frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+......$$
$$\frac{\left(e^h-1\right)}{h}= \frac{1}{1!}+\frac{h}{2!}+\frac{h^2}{3!}+......$$
Thus $$\lim_{h \to 0}\frac{\left(e^h-1\right)}{h}= 1$$
