ksaimouli
  • ksaimouli
linearization
Mathematics
schrodinger
  • schrodinger
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ksaimouli
  • ksaimouli
|dw:1353856476119:dw|
ksaimouli
  • ksaimouli
a=0
anonymous
  • anonymous
What do you have to do?

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ksaimouli
  • ksaimouli
i am gettingy=1x+1 but the answer is -1x+1 nothing but the slope
ksaimouli
  • ksaimouli
find derivative of y to find teh slope
ksaimouli
  • ksaimouli
and then equation of line
anonymous
  • anonymous
\[\huge \frac {d\frac{1}{1+\frac{sinx}{cosx}}} {dx}= \frac {d\frac{1}{1+\frac{sinx}{cosx}}} {d\frac{sinx}{cosx}} \frac {d\frac{sinx}{cosx}} {dx}\]
anonymous
  • anonymous
\[\huge = \frac{-1}{(1+\frac{sinx}{cosx})^2} \frac{1}{cos^2x}\]
ksaimouli
  • ksaimouli
i know tanx=sec^2x
ksaimouli
  • ksaimouli
which is 1/cos^2x
ksaimouli
  • ksaimouli
\[\left(\begin{matrix}-\sec^2(0) \\ 1+\tan^2(0)\end{matrix}\right)\]
ksaimouli
  • ksaimouli
=1???
anonymous
  • anonymous
\[\huge \text{ when } a=0 , \text{ it is }\] \[\huge \frac{-1}{(1-\frac{0}{1})^2}\frac{1}{1^2}=-1\]
anonymous
  • anonymous
\[\huge y=-x+k\] What is k? When x=0, \[\huge y=\frac{1}{1+0}=1\] Inserting \[ \huge (0,1) \] into \[\huge y=-x+k\], we have \[\huge 1=-0+k\] \[ \huge \Rightarrow k=1 \] \[\huge \Rightarrow y=-x+1\]
anonymous
  • anonymous
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIxLygxK3Rhbih4KSkiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjAsImVxIjoiLXgrMSIsImNvbG9yIjoiI0Q0MUUxRSJ9LHsidHlwZSI6MTAwMH1d
anonymous
  • anonymous
Shout if confused

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