Question

What is the derivative of y=[(1+sinx)/(1+cosx)]^2

Answer 1

\[[\frac{f}{g}]'=\frac{ f'g -fg'}{ g^2 }\]

Answer 2

first use chain rule to drop the 2

Answer 3

Use the product rule for this. Well that's what I would use.

Answer 4

Actually, both methods are good.

Answer 5

OMG! Do you want the answer because its really long.

Answer 6

I think I got it, I used the chain rule, and then the quotient rule for the g'(x) portion of the chain rule. This is a very, very ugly problem.

Answer 7

This is what I got. You can use and compare you your answer.

Answer 8

\[2(\frac{\cos x(1+\cos x)+\sin x(1+\sin x)}{(\cos x +1)^2})\]

Answer 9

not that bad

Answer 10

wow theres a lot of ways to do this problem. I got \[(2*(\sin(x) +1)*(\cos(x) +\sin(x) +1)/(\cos(x)+1)^3\]

Answer 11

or (2(5cos(x)+5sin(x)cos(x)+3cos^(2)(x)+3sin(x)cos^(2)(x)+sin(x)+2sin^(2)(x)+sin^(3)(x)))/(1+cos(x))