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

Answer:

$f'(2)=\lim_{h \rightarrow 0} \frac{-1}{4(4+h)}$

Step-by-step explanation:

The definition of derivative is:

$f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$

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

So let's focus on just:

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

We are given $f(x)=\frac{1}{x+2}$.

So $f(x+h)=\frac{1}{(x+h)+2}$ (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:

$\frac{f(2+h)-f(2)}{h}$

$\frac{\frac{1}{(2+h)+2}-\frac{1}{2+2}}{h}$

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

$\frac{\frac{1}{4+h}-\frac{1}{4}}{h}$

Now division by $h$ can be written as multiplication by $\frac{1}{h}$:

$\frac{1}{h}(\frac{1}{4+h}-\frac{1}{4})$

Let's combine the fractions inside the ( ).

I will multiply the first fraction by $1=\frac{4}{4}$.

I will multiply the second fraction by $1=\frac{4+h}{4+h}$.

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

$\frac{1}{h}(\frac{4}{4(4+h)}-\frac{4+h}{4(4+h)})$

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

$\frac{1}{h}(\frac{4-(h+4)}{4(4+h)})$

Let's simplify the numerator in the ( ).

$\frac{1}{h}(\frac{-h}{4(4+h)})$

Now you should see a common factor to cancel. That is we have that $\frac{h}{h}=1$. So we can write that:

$\frac{1}{h}(\frac{-h}{4(4+h)})=\frac{-1}{4(4+h)}$

So the answer we are looking for is:

$f'(2)=\lim_{h \rightarrow 0} \frac{-1}{4(4+h)}$

Answer 2

Answer:

E

Step-by-step 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)]