Question

help me to the derivative
(cos2x + sin2x) / (cos2x - sin2x)

Answer 1

derivative of tan (pi/4+2x)
2sec^2 (pi/4+2x)

Answer 2

I do not understand your solution

Answer 3

\[\frac {\cos2x +\sin2x}{\cos2x-\sin2x}= \frac {1+\tan2x}{1-\tan2x}\]

Answer 4

now do u understand??

Answer 5

I want to resolve it with the derivative and
differentiation rules

Answer 6

hello @maribor76 didn't u understand my approach??

Answer 7

the second post of me have u understood??

Answer 8

\[\frac{\text{Cos}[2 x]+\text{Sin}[2 x]}{\text{Cos}[2 x]-\text{Sin}[2 x]}\text{=}\text{ Csc}\left[\frac{\pi }{4}-2 x\right] \text{Sin}\left[\frac{\pi }{4}+2 x\right] \]The derivative of the RHS is\[2 \text{Cos}\left[\frac{\pi }{4}+2 x\right] \text{Csc}\left[\frac{\pi }{4}-2 x\right]+2 \text{Cot}\left[\frac{\pi }{4}-2 x\right] \text{Csc}\left[\frac{\pi }{4}-2 x\right] \text{Sin}\left[\frac{\pi }{4}+2 x\right] \]The above simplifed is:\[-\frac{4}{\sin (4 x)-1} \]

Answer 9

\[\frac {1+\tan2x}{1-\tan2x}= \tan(\pi/4 +2x)\]\[\frac d{dx} \tan(\frac {\pi}4+2x) = 2\sec^2(\frac {\pi}4+2x)\]

Answer 10

Good work dipankarstudy!

Answer 11

Thank you all now I get it