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.

$$\lim_{h \to 0} \frac{e^x}{\left(\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}\right)} \cdot \frac{\left(e^h-1\right)}{h}$$
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 $$\lim_{h \to 0}\frac{\left(e^h-1\right)}{h}= 1$$
