Derivative of Trig Function | Find y' for (sec (2x))/(1 + tan (2x))

[Jumpy Lee]

Homework Statement

Find y' of the following

Homework Equations

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

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]

 

[cristo]

Try using the identity sec²x=1+tan²x on the last term in the numerator. Then it should simplify to the result you state.

 

[benorin]

You had

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}
=\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}

 

[Jumpy Lee]

thank you cristo and benorin

 

[benorin]

This website can help you with derivatives:

http://library.wolfram.com/webMathematica/Education/WalkD.jsp