Question

Can someone please walk me through this problem:

Answer 1

\[\frac{ \frac{ 1 }{ (x+h)^{2}}-\frac{ 1 }{x ^{2} } }{ h} \]

Answer 2

Glad to help, but would you please include the instructions for this problem first? This problem is very similar to one that you did a few minutes ago; only the sign of the final result is different. You might want to review your previous question.

Answer 3

It is the same problem... I confused myself and got lost and need to start fresh.... instructions are: simplify the fractional expression.

Answer 4

1. Identify the LCD. What is it?

Answer 5

(x+h)^2 * x^2?

Answer 6

yes

Answer 7

You have a couple of options,
1. You can create a common denominator between the fractions.
2. You can multiply through by the LCM of the denominators.

I prefer the second route,\[\large\rm \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\color{royalblue}{\cdot\left(\frac{(x+h)^2\cdot x^2}{(x+h)^2\cdot x^2}\right)}\]
if it's too confusing though, you can do the first.

Answer 8

Ok 1 sec... let me work that out...

Answer 9

\[\frac{ x ^{2-(x+h)^{2}} }{ h(x+h)^{2}x ^{2}}\]

Answer 10

\[\large\rm \frac{x^2-(x+h)^2}{hx^2(x+h)^2}\]Yup, looks good so far.
Now expand out the square in the `numerator`.

Answer 11

x^2-x^2+xh+hx+h^2

Answer 12

x^2+x^2+2hx+h^2

Answer 13

Ok good.
But realize that the subtraction in the numerator is being applied to the `entire expansion`.\[\large\rm x^2-(x^2+xh+hx+h^2)\]

Answer 14

oh shoot ... ok i see what i did

Answer 15

\[\large\rm \frac{x^2-(x^2+2xh+h^2)}{hx^2(x+h)^2}\]This is where we are at so far.

Answer 16

ok... do I exand the (x+h)^2 in the denominator now?

Answer 17

Nahhh I don't think we want to do that.
The whole point of this exercise is to try and cancel out the h in the denominator. Expanding the denominator is just going to make a big mess.

Answer 18

No. Instead, simplify the numerator. x^2 - x^2 = ?

Answer 19

Distribute the negative in the numerator,
and then try to combine like-terms.

Answer 20

After you've done that correctly, you can then cancel out the h from numerator and denominator.

Answer 21

so: \[\frac{x^2(-x^2-2xh-h^2)}{hx^2(x+h)^2} \] ?

Answer 22

After you distribute the negative, drop the brackets,\[\large\rm \frac{x^2-x^2+2xh+h^2}{hx^2(x+h)^2}\]It's important to realize that the operation in front of the brackets shouldn't change from `subtraction` to `multiplication`. That would be weird. That's what it looks like in your expression.

Answer 23

Woahh I didn't distribute the negative, my bad.

Answer 24

\[\large\rm \frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}\]

Answer 25

ok i see

Answer 26

Again: Please simplify the numerator.
Write out your result.
Look for an additional way to reduce the fraction.

Answer 27

\[\frac{-2xh-h^2}{hx^2(x+h)^2}\]

Answer 28

is that right?

Answer 29

Sorry ran off for a moment :)
Looks good so far!
You can factor something out of each term in the numerator.

Answer 30

h(-2x+h)

Answer 31

h(-2x-h)

Answer 32

woops that was a typo... wrote it right on paper

Answer 33

\[\large\rm\frac{h(-2x-h)}{hx^2(x+h)^2}\]From that point, you should see a cancellation available.

Answer 34

\[\frac{(-2x-h)}{x^2(x+h)^2} \]

Answer 35

Yayyyyyy good job \c:/
That's pretty much all that needs to be done.

What type of math are you doing?
In Calculus, we take this difference quotient,
and we let this value h approach 0,
so it's important that it not show up in the denominator as it did before,
that screws things up for us.

Answer 36

Thank you so much for your patience! I am doing pre-calculus... it has been a very long time since I have done math so I am needing to re-learn the basics as I go...

Answer 37

Ahhh I know that game :)
I had to do that a few years back.

Math was always my favorite but I had to start back at college Algebra since I hadn't done math in 10 years. It's actually good retaking some of those simple foundational courses. It really really refreshes your mind.