anonymous
  • anonymous
find d^2y/dx^2 for y=(1-x)/(2-x)
Mathematics
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anonymous
  • anonymous
find d^2y/dx^2 for y=(1-x)/(2-x)
Mathematics
chestercat
  • chestercat
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amistre64
  • amistre64
find the second derivative ... right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
try the quotient rule

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amistre64
  • amistre64
then i would either use a product rule with a ^-1; or the quotient rule
anonymous
  • anonymous
ok thanks so much
amistre64
  • amistre64
youd have to do it twice tho to get to the second one
anonymous
  • anonymous
ok
amistre64
  • amistre64
\[\frac{t}{b}=tb^{-1}\] \[D_x[tb^{-1}]=-tb^{-2}+t'b^{-1}\] \[D_{x}^{2}[tb^{-1}]=D_x[-tb^{-2}+t'b^{-1}]\] \[D_x[-tb^{-2}+b^{-1}]=2tb^{-3}-t'b^{-2}-t'b^{-2}+t''b^{-1}\] witha any luck I kept it all in order lol
amistre64
  • amistre64
t=(1-x) t' = -1 t'' = 0 b=(2-x) \[2(1-x)(2-x)^{-3}+(2-x)^{-2}+(2-x)^{-2}+0(b^{-1})\] \[2(1-x)(2-x)^{-3}+2(2-x)^{-2}\] \[2(2-x)^{-3}\left((1-x)+(2-x)\right)\] \[\frac{2(3)}{(2-x)^{3}}\] the wolf should be able to check that tho
amistre64
  • amistre64
almost kept it straight :) http://www.wolframalpha.com/input/?i=second+derivative+t%28x%29%28b%28x%29%29%5E%28-1%29 2 ------- (x-2)^3 http://www.wolframalpha.com/input/?i=second+derivative+%281-x%29%2F%282-x%29

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