What are the different methods to determine the derivative of $f(x)=\sqrt{1+x}$?

Question

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 $f(x)$?

Answer 1

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}

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$[f(x)]^2=1+x$. So

\begin{align} 2f(x)f'(x)&=1\\ f'(x)&=\frac{1}{2f(x)}\\ &=\frac{1}{2\sqrt{1+x}} \end{align}

Answer 2

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}

Answer 3

Another one.

$$f(x)=\sqrt{1+x}\implies \log(f(x))=\frac 12 \log(1+x)$$ Differentiate both sides $$\frac{f'(x)}{f(x)}=\frac 12 \frac 1{1+x}\implies {f'(x)}=f(x) \times\frac{f'(x)}{f(x)}=\frac 12 \frac 1{1+x} \sqrt{1+x}=\frac{1}{2\sqrt{1+x}}$$

Answer 4

Using division rule:

$$(\sqrt{1+x})'=\left(\frac{1+x}{\sqrt{1+x}}\right)'=\frac{\sqrt{1+x}-(1+x)(\sqrt{1+x})'}{1+x}=\frac{1}{\sqrt{1+x}}-(\sqrt{1+x})' \Rightarrow (\sqrt{1+x})'=\frac{1}{2\sqrt{1+x}}.$$