Question

Which of the following corresponds to the derivative of f(x)=1/(x+2) at x = 2, using the primary definition of a derivative, reduced to its simplest form before taking the limit?

I know the derivative at x=2 is -1/16, but I am confused by the question.

Answer 1

E

Explanation:

[f(2+h) - f(2)]/h

f(2+h) = 1/(2+h+2) = 1/(4+h)

f(2) = 1/(2+2) = 1/4

[1/(4+h) - 1/4] ÷ h

[4 - (4+h)] ÷ [4h(4+h)]

-h ÷ [4h(4+h)]

-1/[4(4+h)]

Answer 2

`f'(2)=\lim_(h \rightarrow 0) (-1)/(4(4+h))`

Explanation:

The definition of derivative is:

`f'(x)=\lim_(h \rightarrow 0)(f(x+h)-f(x))/(h)`

So it asks us not to evaluate the limit part but to simplify the fraction part.

So let's focus on just:

`(f(x+h)-f(x))/(h)`

We are given
`f(x)=(1)/(x+2)`.

So
`f(x+h)=(1)/((x+h)+2)` (I just replaced the
`x`'s with
`(x+h)`'s.)

Now since we want to find it a
`x=2`. I'm going to replace my x's with 2:

So instead we will look at:

`(f(2+h)-f(2))/(h)`

`((1)/((2+h)+2)-(1)/(2+2))/(h)`

Let's simplify some of the addition that we can in the denominators of the mini-fractions:

`((1)/(4+h)-(1)/(4))/(h)`

Now division by
`h` can be written as multiplication by
`(1)/(h)`:

`(1)/(h)((1)/(4+h)-(1)/(4))`

Let's combine the fractions inside the ( ).

I will multiply the first fraction by
`1=(4)/(4)`.

I will multiply the second fraction by
`1=(4+h)/(4+h)`.

We are going to do this so we have the same denominator:

`(1)/(h)((4)/(4(4+h))-(4+h)/(4(4+h)))`

Now we have the same denominator inside the ( ) and can combine those fractions:

`(1)/(h)((4-(h+4))/(4(4+h)))`

Let's simplify the numerator in the ( ).

`(1)/(h)((-h)/(4(4+h)))`

Now you should see a common factor to cancel. That is we have that
`(h)/(h)=1`. So we can write that:

`(1)/(h)((-h)/(4(4+h)))=(-1)/(4(4+h))`

So the answer we are looking for is:

`f'(2)=\lim_(h \rightarrow 0) (-1)/(4(4+h))`