# Thread: Finding a derivative

1. ## Finding a derivative

The problem is this:

Find the derivative of the function $f(x) = \frac{-2}{(x + 1)^2}$ using a difference quotient. Hint: leave the denominator in factored form.

Also, I should mention, the only way my class has learned to solve for this so far is using the equation: $f'(x) = \lim h \rightarrow 0 \frac{f(x + h) - f(x)}{h}$

So, I believe I've solved the problem, but the way in which I did it seems incorrect, even though it yields the correct answer (or maybe there's a better way...I just feel like my way isn't what would be preferred!)

Here's what I did:

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

2.) $f'(x) = \lim h \rightarrow 0 \frac{ \frac{-2}{(x + h + 1)^2} - \frac{-2}{(x + 1)^2} }{h}$

3.) I then found a common denominator and combined the top. I also made the h on the bottom into $\frac {1}{h}$ to make it a little easier to write. That gave me this:

$f'(x) = \lim h \rightarrow 0 (\frac{1}{h})
(\frac{2(x + h + 1)^2 - (2(x + 1)^2)}{(x + 1)^2 (x + h + 1)^2})
$

4.) For the sake of not writing out all the algebra, I multiplied the entire top part through and ended up with this:

$f'(x) = \lim h \rightarrow 0 (\frac{1}{h}) (\frac{2h^2 + 4hx + 4h}{(x + 1)^2(x + h + 1)^2})
$

5.) Then I factored out an "h" from the top and canceled it with the 1/h, getting this:

$f'(x) = \lim h \rightarrow 0 (\frac{2h + 4x + 4}{(x + 1)^2(x + h + 1)^2})
$

6.) Now here's where I think what I've done might not be considered correct. I got stuck once I reached this point, so I attempted to just take the limit of all of that at once like this (sorry for how long this is!):

$f'(x) =\frac{
\lim h \rightarrow 0 (2h) + \lim h \rightarrow 0 (4x) + \lim h \rightarrow 0 (4)}
{[\lim h \rightarrow 0 (x + 1)]^2[\lim h \rightarrow 0 (x) + \lim h \rightarrow 0 (h) + \lim h \rightarrow 0 (1)]^2}
$

Which, after figuring all of that out gave me this:
$
\frac{2(0) + 4x + 4}{(x+1)^2(x + 0 + 1)(x + 0 + 1)}$

If you simplify that you get

$\frac{4x + 4}{(x + 1)^2 (x + 1)^2}$

Factor out a 4 from the top, cancel, and you get a final answer of

$\frac {4}{(x + 1)^3}$

Now, that's the right answer but...am I getting to it correctly? That's my question.

Any help at all would be much appreciated! I know it's a really long problem.

2. Originally Posted by lysserloo
The problem is this:

Find the derivative of the function $f(x) = \frac{-2}{(x + 1)^2}$ using a difference quotient. Hint: leave the denominator in factored form.

Also, I should mention, the only way my class has learned to solve for this so far is using the equation: $f'(x) = \lim h \rightarrow 0 \frac{f(x + h) - f(x)}{h}$

So, I believe I've solved the problem, but the way in which I did it seems incorrect, even though it yields the correct answer (or maybe there's a better way...I just feel like my way isn't what would be preferred!)

Here's what I did:

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

2.) $f'(x) = \lim h \rightarrow 0 \frac{ \frac{-2}{(x + h + 1)^2} - \frac{-2}{(x + 1)^2} }{h}$

3.) I then found a common denominator and combined the top. I also made the h on the bottom into $\frac {1}{h}$ to make it a little easier to write. That gave me this:

$f'(x) = \lim h \rightarrow 0 (\frac{1}{h})
(\frac{2(x + h + 1)^2 - (2(x + 1)^2)}{(x + 1)^2 (x + h + 1)^2})
$

4.) For the sake of not writing out all the algebra, I multiplied the entire top part through and ended up with this:

$f'(x) = \lim h \rightarrow 0 (\frac{1}{h}) (\frac{2h^2 + 4hx + 4h}{(x + 1)^2(x + h + 1)^2})
$

5.) Then I factored out an "h" from the top and canceled it with the 1/h, getting this:

$f'(x) = \lim h \rightarrow 0 (\frac{2h + 4x + 4}{(x + 1)^2(x + h + 1)^2})
$

6.) Now here's where I think what I've done might not be considered correct. I got stuck once I reached this point, so I attempted to just take the limit of all of that at once like this (sorry for how long this is!):

$f'(x) =\frac{
\lim h \rightarrow 0 (2h) + \lim h \rightarrow 0 (4x) + \lim h \rightarrow 0 (4)}
{[\lim h \rightarrow 0 (x + 1)]^2[\lim h \rightarrow 0 (x) + \lim h \rightarrow 0 (h) + \lim h \rightarrow 0 (1)]^2}
$

Which, after figuring all of that out gave me this:
$
\frac{2(0) + 4x + 4}{(x+1)^2(x + 0 + 1)(x + 0 + 1)}$

If you simplify that you get

$\frac{4x + 4}{(x + 1)^2 (x + 1)^2}$

Factor out a 4 from the top, cancel, and you get a final answer of

$\frac {4}{(x + 1)^3}$

Now, that's the right answer but...am I getting to it correctly? That's my question.

Any help at all would be much appreciated! I know it's a really long problem.

you solution is correct but in the 6th step there is no need to separate the limit to all just sub h=0 just

and write the limit like this \lim_{h\rightarrow 0} to be like this $\lim_{h\rightarrow 0}$ or like this lim_{h\rightarrow 0} to be like $lim_{h\rightarrow 0}$