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Deriv of trig function

  • Thread starter Jumpy Lee
  • Start date
20
0
1. Homework Statement

Find y' of the following

2. Homework Equations

y = (sec (2x))/(1 + tan (2x))

3. The Attempt at a Solution

y' = [(1 + tan (2x))(2 sec (2x) tan (2x)) - (sec (2x))(2 (sec (2x))^2)]/[(1 + tan (2x))^2]


i can not quite figure out how it reduces to:

y' = [(2 sec (2x))(tan (2x) - 1)]/[(1 + tan (2x))^2] :grumpy:
 

Answers and Replies

cristo
Staff Emeritus
Science Advisor
8,056
72
Try using the identity sec2x=1+tan2x on the last term in the numerator. Then it should simplify to the result you state.
 
Last edited:
benorin
Homework Helper
1,067
14
You had

[tex]y^\prime =\frac{(1 + \tan 2x)(2\sec 2x \tan 2x) - (\sec 2x)(2 \sec^{2} 2x)}{(1 + \tan 2x)^2}=\frac{2\sec 2x \left[(1 + \tan 2x)\tan 2x - \sec^{2} 2x\right]}{(1 + \tan 2x)^2}[/tex]
[tex]=\frac{2\sec 2x \left[\tan 2x + \tan^{2} 2x - (1 + \tan^{2} 2x)\right]}{(1 + \tan 2x)^2}=\frac{2\sec 2x (\tan 2x - 1)}{(1 + \tan 2x)^2}[/tex]
 
20
0
thank you cristo and benorin
 

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