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ksaimouli
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|dw:1353856476119:dw|
ksaimouli
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a=0
henpen
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What do you have to do?
ksaimouli
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i am gettingy=1x+1 but the answer is -1x+1 nothing but the slope
ksaimouli
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find derivative of y to find teh slope
ksaimouli
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and then equation of line
henpen
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\[\huge \frac {d\frac{1}{1+\frac{sinx}{cosx}}} {dx}= \frac {d\frac{1}{1+\frac{sinx}{cosx}}} {d\frac{sinx}{cosx}} \frac {d\frac{sinx}{cosx}} {dx}\]
henpen
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\[\huge = \frac{-1}{(1+\frac{sinx}{cosx})^2} \frac{1}{cos^2x}\]
ksaimouli
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i know tanx=sec^2x
ksaimouli
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which is 1/cos^2x
ksaimouli
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\[\left(\begin{matrix}-\sec^2(0) \\ 1+\tan^2(0)\end{matrix}\right)\]
ksaimouli
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=1???
henpen
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\[\huge \text{ when } a=0 , \text{ it is }\]
\[\huge \frac{-1}{(1-\frac{0}{1})^2}\frac{1}{1^2}=-1\]
henpen
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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 \[\huge 1=-0+k\]
\[ \huge \Rightarrow k=1 \]
\[\huge \Rightarrow y=-x+1\]
henpen
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Shout if confused