Question

First derivative of x/x^2+1
I am confused with the algebraic simplification.
Should I use quotient rule?

Answer 1

I'm assuming you mean x/(x^2+1)? Make sure to use brackets!

Yes you should use quotient rule. f=x, g=x^2+1

Answer 2

ok I get 1(x^2+1)-x(2x)/(x^2+1)^2
???

Answer 3

Brackets!

(1(x^2+1)-x(2x))/(x^2+1)^2

Answer 4

ok this next part I can't do correctly

Answer 5

Simplifying it? Just expand the top of the fraction and combine common terms.

Answer 6

I can't do it Ive tried believe me lol

Answer 7

In my eyes the 1 distributes into the (x^2+1)?

Answer 8

1(x^2+1)-x(2x)
=x^2+1-2x^2
=-x^2+1

Answer 9

Yes, the 1 distributes.

Answer 10

Ok so the whole thing reduces top and bottom to -x^2+1 right?

Answer 11

No, just the top. The bottom doesn't change.

Answer 12

x= +/- 1
?

Answer 13

Are you trying to solve for x when it is equals 0 or something now?

Answer 14

yes

Answer 15

one of the x^2+1 cancels?

Answer 16

So now you have (-x^2+1)/(x^2+1)^2 right?

To solve for when it equals 0, you only need to solve for when the NUMERATOR equals 0. The denominator can be ignored.

so solve for

-x^2+1=0

You should get x=+-1, so yeah you're right.

Answer 17

oh crap I think that is the first time I heard its numerator only. Thanks alot!

Answer 18

Think about it this way:
\[\frac{ -x^2+1 }{ (x^2+1)^2 }=0\]Multiply both sides by (x^2+1)^2. You get
\[-x^2+1=0\]So you can always multiply both sides by the denominator and it just goes away.

Answer 19

Ahhhh ok that makes sense