Question

Find the derivative of the function at the given number.

.$f\left(x\right)=\frac{1}{x+1},at2$

Short answer

The derivative of the function at2 is.$-\frac{1}{9}$

Step-by-step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=\frac{1}{x+1},at\text{2}$

Step 2. Formula used.

The derivative of a function f at a number,denoted by$f\text{'}\left(a\right),$is

.$f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$

if this limit exists.

Step 3. Finding the slope of the tangent line at a given point.

According to the definition of a derivative, with,$a=2$ we have

$\begin{array}{c}f\text{'}\left(2\right)=\underset{h\to 0}{\lim }\frac{f(2+h)-f\left(2\right)}{h}Definitionoff\text{'}\left(2\right)\\ =\underset{h\to 0}{\lim }\frac{\left[\frac{1}{(2+h)+1}\right]-\left[\frac{1}{2+1}\right]}{h}f\left(x\right)=\frac{1}{x+1}\\ =\underset{h\to 0}{\lim }\frac{\frac{1}{h+3}-\frac{1}{3}}{h}Expand\\ =\underset{h\to 0}{\lim }\frac{\frac{-h}{3(h+3)}}{h}\\ =\underset{h\to 0}{\lim }\frac{-h}{3h(h+3)}Simplify\\ =\underset{h\to 0}{\lim }\frac{-1}{3(h+3)}Cancelh\\ =-\frac{1}{9}\end{array}$

Therefore, the derivative of the function at2 is $-\frac{1}{9}$.