What are the different methods to determine the derivative of $f(x)=\sqrt{1+x}$? The methods that I can think of are:
1) Chain rule
2) Binomial series of $f(x)$
3) Through the formula $f\prime(g(x))=\frac{1}{g\prime (x)}$ at a particular point
What are the other methods to determine the derivative of $f(x)$?
 A: From first principles.
\begin{align}
f'(x)&=\lim_{h\to0}\frac{\sqrt{1+x+h}-\sqrt{1+x}}{h}\\
&=\lim_{h\to0}\frac{(\sqrt{1+x+h}-\sqrt{1+x})(\sqrt{1+x+h}+\sqrt{1+x})}{h(\sqrt{1+x+h}+\sqrt{1+x})}\\
&=\lim_{h\to0}\frac{1+x+h-1-x}{h(\sqrt{1+x+h}+\sqrt{1+x})}\\
&=\lim_{h\to0}\frac{1}{\sqrt{1+x+h}+\sqrt{1+x}}\\
&=\frac{1}{2\sqrt{1+x}}
\end{align}

$[f(x)]^2=1+x$. So
\begin{align}
2f(x)f'(x)&=1\\
f'(x)&=\frac{1}{2f(x)}\\
&=\frac{1}{2\sqrt{1+x}}
\end{align}
A: Another one.
$$f(x)=\sqrt{1+x}\implies \log(f(x))=\frac 12 \log(1+x)$$ Differentiate both sides $$\frac{f'(x)}{f(x)}=\frac 12 \frac 1{1+x}\implies {f'(x)}=f(x) \times\frac{f'(x)}{f(x)}=\frac 12  \frac 1{1+x} \sqrt{1+x}=\frac{1}{2\sqrt{1+x}}$$
A: Using division rule:
$$(\sqrt{1+x})'=\left(\frac{1+x}{\sqrt{1+x}}\right)'=\frac{\sqrt{1+x}-(1+x)(\sqrt{1+x})'}{1+x}=\frac{1}{\sqrt{1+x}}-(\sqrt{1+x})' \Rightarrow (\sqrt{1+x})'=\frac{1}{2\sqrt{1+x}}.$$
A: Using logarithmic differentiation:
\begin{align}
f'(x) &= f(x) \cdot \frac{d}{dx} \ln f(x)
= \sqrt{1+x} \cdot \frac{d}{dx} \ln \sqrt{1+x}\\
&= \sqrt{1+x} \cdot\frac{1}{2} \frac{d}{dx} \ln(1+x)
= \sqrt{1+x} \cdot \frac{1}{2} \frac{1}{1+x}\\
&= \frac{1}{2\sqrt{1+x}}
\end{align}
