Deriving the Double Derivative of tan(x)

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Homework Help Overview

The discussion revolves around deriving the double derivative of the function y = tan(x). Participants are exploring the differentiation process and simplification of the resulting expressions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial derivative calculation and question the appropriate methods for finding the second derivative, considering the use of the quotient and chain rules. There is also an exploration of how to simplify the expression for the second derivative.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's attempts. Some guidance has been offered regarding differentiation techniques, and there is a recognition of the correct form of the second derivative, although simplification remains a topic of exploration.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the resources they can reference or the methods they can employ.

_wolfgang_
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Homework Statement


the question asked to prove that the double derivaitve of y=tan(x) is...
2y(1+y^2)
eg. 2tanx(1+tan^2(x))



Homework Equations





The Attempt at a Solution



I was able to get the first derivative ( i think)
y=tan(x)
=(sin(x))/(cos(x))

dy/dx=(cos(x)cos(x)-(sin(x)(-sin(x))))/cos^2(x)
=(cos^2(x)+sin^2(x))/cos^2(x)
=1/cos^2(x)

from here i am not to sure how to get the second derivative...
 
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So far so good, just differentiate cos-2x.
 
So would i use the quotient rule to do that or something else?? like the chain rule??
 
ok iv done that i get
2sin(x)/cos^3(x)
now iv got the double derivative i can't see how to simplify it to get 2tanx(1+tan^2(x))
 
All I can tell is that

[tex]\frac {2 \sin x} {\cos^3 x} = \frac {2 \tan x } {\cos^2 x}[/tex]

is a correct second derivative.
 
ok thanks for the help i should be able to get it from here
 

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