Question

I know this may sound like a dumb question but I am having a memory lapse...how do you find the derivative of (1)/(2X)

Answer 1

\[= (2x)^{-1}\]

use the chain rule and power rule

Answer 2

I am just kidding the problem was (1)/(x^2-1)

Answer 3

or\[=\frac{ 1 }{ 2 }x ^{-1}\]
just the power rule

Answer 4

ok thank you very much:)

Answer 5

then rewrite like i did in the first post and use the chain rule

Answer 6

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

Answer 7

right so then it would be -1(x^2-1)^-2 times (2x)

Answer 8

\[\frac{ d }{ dx }(x^{2}-1)^{-1} = -(2x)(x^{2}-1)^{-2}=-\frac{ 2x }{ (x^{2}-1)^{2} }\]

Answer 9

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

Answer 10

you got it!

Answer 11

ok thanks \[\frac{ 2x }{ x ^{4}-2x+1 }\]

Answer 12

oh with a negative in front?

Answer 13

why did you multiply that out? it really isn't necessary and could be a pain if you were to take further derivatives.

Answer 14

because i am actually using it to find the points of the graph here the tangent line is horizontal so i have to set it equal to 0 and solve for the point

Answer 15

you'll end up factoring it again. the bottom is 0 when x = 1 or -1. however, the tangent will not be horizontal there but rather it will be vertical.
set the top = 0 and solve to find horizontal tangents.

Answer 16

ohhh lol you're right

Answer 17

sooo im confused as to what to do after setting it equal to 0

Answer 18

solve for x. that's where you'll get a horizontal tangent.

Answer 19

i know that but how

Answer 20

\[-\frac{ 2x }{ (x^{2}-1)^{2} }=0 \Rightarrow 2x = 0 \text{ solve}\]

Answer 21

what?

Answer 22

ohh so set both the top and the bottom equal to 0?

Answer 23

\[2x=0 and (x ^{2}-1)^{2}=0\]