Polynomial Simplification

**dbakeg00** · March 4th 2010, 11:26 AM

First time poster here!

I am reading through some calculus proofs on derivatives and I'm a little rusty on some of my algebra. Would someone mind helping me figure out the steps to go from #1 to #2?

#1 $(x+dx)^{-2}$

#2 $x^{-2}*(1+\frac{dx}{x})^{-2}$

I would like to politely request a step by step answer if you wouldn't mind because i don't remember much about this from Algebra. You will not insult my intelligence. Thanks in advance for your help.

-db

**HallsofIvy** · March 4th 2010, 11:50 AM

Originally Posted by dbakeg00

First time poster here!

I am reading through some calculus proofs on derivatives and I'm a little rusty on some of my algebra. Would someone mind helping me figure out the steps to go from #1 to #2?

#1 $(x+dx)^{-2}$

#2 $x^{-2}*(1+\frac{dx}{x})^{-2}$

To $\frac{1}{(x+ dx)^2}= \frac{1}{x^2+ 2x dx+ (dx)^2}$

Start buy factoring an "x" out of (x+ dx): since $dx= x\frac{dx}{x}$ that is $x(1+ \frac{dx}{x})$

Now take both to the -2 power: $x^{-2}(1+ \frac{dx}{x})^{-2}$

I would like to politely request a step by step answer if you wouldn't mind because i don't remember much about this from Algebra. You will not insult my intelligence. Thanks in advance for your help.

-db

**dbakeg00** · March 4th 2010, 11:54 AM

Thank you.

I guess I never knew that $ dx= x\frac{dx}{x} $ ...that must be why I couldn't see how you got from the first expression to the second. That helped a lot. Thank you again.

db

**dbakeg00** · March 4th 2010, 01:06 PM

Anyone care to explain why $ dx= x\frac{dx}{x} $ ...I'm not sure I remember that

thanks,
db

**HallsofIvy** · March 5th 2010, 03:13 AM

You cancel the two "x"s. You learned that in the third grade.

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