Deriving the Double Derivative of tan(x)

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The discussion focuses on deriving the double derivative of y=tan(x), which is stated to be 2y(1+y^2) or 2tan(x)(1+tan^2(x)). The first derivative is correctly calculated as dy/dx = 1/cos^2(x). The participant expresses uncertainty about finding the second derivative and considers using either the quotient or chain rule for differentiation. After some calculations, they arrive at the second derivative as 2sin(x)/cos^3(x) and recognize a relationship to 2tan(x)/cos^2(x). The participant is confident they can simplify it to the desired form with further assistance.
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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

\frac {2 \sin x} {\cos^3 x} = \frac {2 \tan x } {\cos^2 x}

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