Deriving the Double Derivative of tan(x)

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SUMMARY

The double derivative of y=tan(x) is proven to be 2tan(x)(1+tan²(x)). The first derivative is calculated as dy/dx = 1/cos²(x), derived using the quotient rule. The second derivative is found to be 2sin(x)/cos³(x), which simplifies to the desired form using the identity tan(x) = sin(x)/cos(x). This discussion clarifies the differentiation process and the application of trigonometric identities in calculus.

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  • Understanding of calculus, specifically differentiation techniques.
  • Familiarity with trigonometric functions, particularly tan(x).
  • Knowledge of the quotient rule and chain rule for derivatives.
  • Ability to manipulate trigonometric identities.
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  • Study the application of the quotient rule in calculus.
  • Learn about trigonometric identities and their simplifications.
  • Explore higher-order derivatives of trigonometric functions.
  • Practice problems involving the differentiation of composite functions.
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Students studying calculus, particularly those focusing on differentiation of trigonometric functions, and educators looking for examples of derivative proofs.

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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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