anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[= (2x)^{-1}\] use the chain rule and power rule
anonymous
  • anonymous
I am just kidding the problem was (1)/(x^2-1)
anonymous
  • anonymous
or\[=\frac{ 1 }{ 2 }x ^{-1}\] just the power rule

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anonymous
  • anonymous
ok thank you very much:)
anonymous
  • anonymous
then rewrite like i did in the first post and use the chain rule
anonymous
  • anonymous
\[\frac{ 1 }{ x^{2}-1 }=(x^{2}-1)^{-1}\]
anonymous
  • anonymous
right so then it would be -1(x^2-1)^-2 times (2x)
anonymous
  • anonymous
\[\frac{ d }{ dx }(x^{2}-1)^{-1} = -(2x)(x^{2}-1)^{-2}=-\frac{ 2x }{ (x^{2}-1)^{2} }\]
anonymous
  • anonymous
\[-1(x^{2}-1)^{-2}(2x)\]
anonymous
  • anonymous
you got it!
anonymous
  • anonymous
ok thanks \[\frac{ 2x }{ x ^{4}-2x+1 }\]
anonymous
  • anonymous
oh with a negative in front?
anonymous
  • anonymous
why did you multiply that out? it really isn't necessary and could be a pain if you were to take further derivatives.
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
ohhh lol you're right
anonymous
  • anonymous
sooo im confused as to what to do after setting it equal to 0
anonymous
  • anonymous
solve for x. that's where you'll get a horizontal tangent.
anonymous
  • anonymous
i know that but how
anonymous
  • anonymous
\[-\frac{ 2x }{ (x^{2}-1)^{2} }=0 \Rightarrow 2x = 0 \text{ solve}\]
anonymous
  • anonymous
what?
anonymous
  • anonymous
ohh so set both the top and the bottom equal to 0?
anonymous
  • anonymous
\[2x=0 and (x ^{2}-1)^{2}=0\]
anonymous
  • anonymous
|dw:1378862327677:dw|
anonymous
  • anonymous
this is a graph of the original function, not the derivative
anonymous
  • anonymous
oook o_O
anonymous
  • anonymous
a picture is worth 1000 words
anonymous
  • anonymous
lol well then i don't consider graphs pictures
anonymous
  • anonymous
maybe you should
anonymous
  • anonymous
:P ew
anonymous
  • anonymous
have you seen a graph of the mandelbrot set? http://www.youtube.com/watch?v=0jGaio87u3A have a look
anonymous
  • anonymous
wow oh my goodness
anonymous
  • anonymous
yes that was but not ones like 1 over x squared minus 1 to the second
anonymous
  • anonymous
http://www.maths.tcd.ie/~nryan/mandelbrot/how.html well, it's not that far off...

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