Asked by Beth
Find the derivative of y=(sinx/(1+cosx))^2.
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
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
Answers
Answered by
Damon
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.
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.
Answered by
Beth
I think you did the quotient rule wrong. Should be minus not plus
Answered by
Beth
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?
Answered by
Beth
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?
Answered by
Steve
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.