Find the derivative of y=(sinx/(1+cosx))^2.

Question

I used the chain rule, but the second part involved the quotient rule which gives a very complicated expression that u do not know how to simplify. Could you please give me a detailed step by step solution for the simplifying part?

Thanks

Answer 1

2(sinx/(1+cosx))d/dx (sinx/(1+cosx))

let y = sinx/(1+cosx)
dy/dx=[(1+cos)cos + sin cos ]/(1+cos)^2

= cos[ 1+cos + sin]/ (1+cos)^2

so put that back in

2(sin/(1+cos))*cos[1+cos+sin]/ (1+cos)^2

2 sin cos(1+cos+sin)/(1+cos)^3

I do not see any simplification either.

Answer 2

I think you did the quotient rule wrong. Should be minus not plus

Answer 3

Wait sorry, you just expanded. But shouldn't it be sin times sin? Not sin times cos? Since derivative of the 1+cos is -sin and the numerator is a sin?

Answer 4

So the answer is supposed to be sin(2x)/(1+cosx)^3 but I was wondering for the numerator why the distributive law doesn't apply when we would multiply the derivative of the outside with the inside?

Answer 5

I think you've both gone stray. See

http://www.wolframalpha.com/input/?i=derivative+%28sinx%2F%281%2Bcosx%29%29

and

http://www.wolframalpha.com/input/?i=derivative+%28sinx%2F%281%2Bcosx%29%29^2