# If f(x)=1/(x+1) how do you evaluate (f(x+h)-f(x))/h?

Aug 2, 2016

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

#### Explanation:

As $f \left(x\right) = \frac{1}{x + 1}$, f(x+h)=1/(x+h+1)# and

$\frac{f \left(x + h\right) - f \left(x\right)}{h}$

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

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

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

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

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

Aug 2, 2016

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

#### Explanation:

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

$\Rightarrow f \left(x + h\right) - f \left(x\right) = \frac{1}{x + h + 1} - \frac{1}{x + 1} = \frac{x + 1 - x - h - 1}{\left(x + 1\right) \left(x + h + 1\right)}$.

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

$\Rightarrow \frac{f \left(x + h\right) - f \left(x\right)}{h} = - \frac{\cancel{h}}{\cancel{h} \left(x + 1\right) \left(x + h + 1\right)}$

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