## ksaimouli 3 years ago linearization

1. ksaimouli

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2. ksaimouli

a=0

3. henpen

What do you have to do?

4. ksaimouli

i am gettingy=1x+1 but the answer is -1x+1 nothing but the slope

5. ksaimouli

find derivative of y to find teh slope

6. ksaimouli

and then equation of line

7. henpen

$\huge \frac {d\frac{1}{1+\frac{sinx}{cosx}}} {dx}= \frac {d\frac{1}{1+\frac{sinx}{cosx}}} {d\frac{sinx}{cosx}} \frac {d\frac{sinx}{cosx}} {dx}$

8. henpen

$\huge = \frac{-1}{(1+\frac{sinx}{cosx})^2} \frac{1}{cos^2x}$

9. ksaimouli

i know tanx=sec^2x

10. ksaimouli

which is 1/cos^2x

11. ksaimouli

$\left(\begin{matrix}-\sec^2(0) \\ 1+\tan^2(0)\end{matrix}\right)$

12. ksaimouli

=1???

13. henpen

$\huge \text{ when } a=0 , \text{ it is }$ $\huge \frac{-1}{(1-\frac{0}{1})^2}\frac{1}{1^2}=-1$

14. henpen

$\huge y=-x+k$ 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
16. henpen

Shout if confused