Differentiation?

Question

How do you differentiate: (sec2x)/(tan2x+1)

I realize it involves the quotient rule, but no matter which way I try to solve it, I never reach [2(sec2x)(tan2x-1)]/(1+ tan2x)^2

Thanks!

Answer 1

sec(2x)
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tan(2x) + 1

Quotient rule combined with chain rule gives you:

[(tan(2x) + 1)(sec(2x)tan(2x))(2) - sec(2x)(sec²(2x))(2)]
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(tan(2x) + 1)²

2sec(2x)[(tan(2x) + 1)(tan(2x)) - (sec²(2x))]
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(tan(2x) + 1)²

Factor out 2sec(2x):

2sec(2x)[tan²(2x) + tan(2x) - sec²(2x)]
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(tan(2x) + 1)²

Then you need to use the identity that tan² = sec² - 1:

2sec(2x)[sec²(2x) - 1 + tan(2x) - sec²(2x)]
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(tan(2x) + 1)²

And you're left with what you want:

2sec(2x)[tan(2x) - 1]
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(tan(2x) + 1)²

Answer 2

The derivative of sec2x is 2sec2x(tan2x)
The derivative of tan2x +1 is 2(sec2x)^2

Now let's apply the quotient rule.
[(sec2x)/(tan2x+1)]'
= [(tan2x+1)(2sec2x(tan2x)) - (sec2x)2(sec2x)^2] / (tan2x+1)^2
Simplify the top.
(tan2x+1)(2sec2x(tan2x)) - (sec2x)2(sec2x)^2
= 2(tan2x)^2(sec2x) + 2sec2x(tan2x) - 2(sec2x)^3
= 2(sec2x)[(tan2x)^2 + tan2x - (sec2x)^2]

Recall the property that (sinx)^2 + (cosx)^2 = 1
Divide everything by (cosx)^2
This gives you the equation
(tanx)^2 + 1 = (secx)^2
Rearranging this equation gives you
(tanx)^2 - (secx)^2 = -1

= 2(sec2x)[(tan2x)^2 + tan2x - (sec2x)^2]
= = 2(sec2x)[(tan2x)^2 - (sec2x)^2 + tan2x]
= 2(secx)[-1 + tan2x]
= 2(secx)(tan2x+1)

Answer: [2(sec2x)(tan2x-1)]/(1+ tan2x)^2