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ksaimouli

  • 3 years ago

linearization

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  1. ksaimouli
    • 3 years ago
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    |dw:1353856476119:dw|

  2. ksaimouli
    • 3 years ago
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    a=0

  3. henpen
    • 3 years ago
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    What do you have to do?

  4. ksaimouli
    • 3 years ago
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    i am gettingy=1x+1 but the answer is -1x+1 nothing but the slope

  5. ksaimouli
    • 3 years ago
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    find derivative of y to find teh slope

  6. ksaimouli
    • 3 years ago
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    and then equation of line

  7. henpen
    • 3 years ago
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    \[\huge \frac {d\frac{1}{1+\frac{sinx}{cosx}}} {dx}= \frac {d\frac{1}{1+\frac{sinx}{cosx}}} {d\frac{sinx}{cosx}} \frac {d\frac{sinx}{cosx}} {dx}\]

  8. henpen
    • 3 years ago
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    \[\huge = \frac{-1}{(1+\frac{sinx}{cosx})^2} \frac{1}{cos^2x}\]

  9. ksaimouli
    • 3 years ago
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    i know tanx=sec^2x

  10. ksaimouli
    • 3 years ago
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    which is 1/cos^2x

  11. ksaimouli
    • 3 years ago
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    \[\left(\begin{matrix}-\sec^2(0) \\ 1+\tan^2(0)\end{matrix}\right)\]

  12. ksaimouli
    • 3 years ago
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    =1???

  13. henpen
    • 3 years ago
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    \[\huge \text{ when } a=0 , \text{ it is }\] \[\huge \frac{-1}{(1-\frac{0}{1})^2}\frac{1}{1^2}=-1\]

  14. henpen
    • 3 years ago
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    \[\huge y=-x+k\] What is k? When x=0, \[\huge y=\frac{1}{1+0}=1\] Inserting \[ \huge (0,1) \] into \[\huge y=-x+k\], we have \[\huge 1=-0+k\] \[ \huge \Rightarrow k=1 \] \[\huge \Rightarrow y=-x+1\]

  15. henpen
    • 3 years ago
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    Shout if confused

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