# Simplify \frac{f(x + h) - f(x)}{h} where f(x) = \frac{1}{x + 1}.

## Question:

Simplify {eq}\displaystyle\;\frac{f(x + h) - f(x)}{h}\; {/eq} where {eq}\displaystyle\;f(x) = \frac{1}{x + 1} {/eq}.

## Derivatives:

Derivatives can be calculated through a variety of formulas depending on the function being differentiated. In order to get a generalized derivative formula for any function {eq}\displaystyle \rm f(x) {/eq}, we use:

{eq}\displaystyle \rm \frac{df}{dx} = \lim_\limits{h\ \to\ 0} \dfrac{f(x+h) - f(x)}{h} {/eq}

## Answer and Explanation:

Given the function:

{eq}\displaystyle \rm f(x) = \frac{1}{x+1} {/eq}

We can substitute this into our general formula:

{eq}\displaystyle \rm \frac{f(x+h) - f(x)}{h} {/eq}

Direct substitution gives us:

{eq}\displaystyle \rm \frac{f(x+h) - f(x)}{h} = \dfrac{\dfrac{1}{x+h+1} - \dfrac{1}{x+1}}{h} {/eq}

By the rules of complex fractions, the *h* term can be transferred to the numerator:

{eq}\displaystyle \rm \frac{f(x+h) - f(x)}{h} = \frac{1}{h}\left(\dfrac{1}{x+h+1} - \dfrac{1}{x+1}\right) {/eq}

Let us now use the least common denominator method. We can multiply:

{eq}\displaystyle \rm \frac{f(x+h) - f(x)}{h} = \frac{1}{h}\left(\dfrac{1}{x+h+1} - \dfrac{1}{x+1}\right) \Big[\frac{(x+h+1)(x+1)}{(x+h+1)(x+1)}\Big] {/eq}

We distribute these multipliers to get:

{eq}\displaystyle \rm \frac{f(x+h) - f(x)}{h} = \frac{1}{h}\left(\dfrac{x+1 - x - h - 1}{(x+h+1)(x+1)}\right) {/eq}

We subtract everything that we can subtract at the numerator:

{eq}\displaystyle \rm \frac{f(x+h) - f(x)}{h} = \frac{1}{h}\left(\dfrac{- h }{(x+h+1)(x+1)}\right) {/eq}

Now we can also cancel out the *h* terms to thus obtain:

{eq}\displaystyle \rm \boxed{\rm \frac{f(x+h) - f(x)}{h} = \dfrac{-1}{(x+h+1)(x+1)}} {/eq}

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