Find from first principles, the derivative of 11+x

To find the derivative of $f(x)=\frac{1}{1+x}$ from first principles, we use the definition of the derivative:

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

Step 1: Write $f(x+h)$

Given $f(x)=\frac{1}{1+x}$, we have:

$$f(x+h)=\frac{1}{1+(x+h)}=\frac{1}{1+x+h}$$

Step 2: Compute $f(x+h)-f(x)$

$$f(x+h)-f(x)=\frac{1}{1+x+h}-\frac{1}{1+x}$$

To combine these fractions, find a common denominator:

$$=\frac{(1+x)-(1+x+h)}{(1+x+h)(1+x)}$$

Simplify the numerator:

$$=\frac{1+x-1-x-h}{(1+x+h)(1+x)}=\frac{-h}{(1+x+h)(1+x)}$$

Step 3: Form the difference quotient

$$\frac{f(x+h)-f(x)}{h}=\frac{\frac{-h}{(1+x+h)(1+x)}}{h}=\frac{-1}{(1+x+h)(1+x)}$$

Step 4: Take the limit as $h\to 0$

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{-1}{(1+x+h)(1+x)}=\frac{-1}{(1+x{)}^2}$$

Final Answer

The derivative of $\frac{1}{1+x}$ from first principles is:

$$\boxed{-\frac{1}{(1+x{)}^2}}$$