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
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.
I think you did the quotient rule wrong. Should be minus not plus
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?
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?
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
To find the derivative of the function y=(sinx/(1+cosx))^2, you correctly applied the chain rule. However, in order to simplify the expression, we need to use some trigonometric identities. Here's a step-by-step solution:
Step 1: Apply the chain rule to differentiate the outer function and keep the inner function unchanged. The outer function is f(x) = u^2, where u = sinx/(1+cosx). The derivative of f(x) with respect to u is 2u.
Step 2: Differentiate the inner function, u = sinx/(1+cosx), using the quotient rule. The quotient rule states that if u = f(x)/g(x), then u' = (f'(x)g(x) - f(x)g'(x))/[g(x)]^2.
Let's compute the derivatives:
f(x) = sinx, so f'(x) = cosx.
g(x) = 1 + cosx, so g'(x) = -sinx (derivative of cosx is -sinx).
Now we can use the quotient rule:
u' = ((cosx)(1+cosx) - (sinx)(-sinx))/[(1+cosx)]^2
= (cosx + cos^2(x) + sin^2(x))/[(1+cosx)]^2
= (cosx + 1)/[(1+cosx)]^2
= 1/[(1+cosx)]
Step 3: Multiply the result from step 2 with the derivative of the outer function (2u). Recall that u = sinx/(1+cosx).
So, the derivative of y=(sinx/(1+cosx))^2 is:
dy/dx = 2u * (1/[(1+cosx)])
= 2(sin(x)/(1+cos(x))) * (1/[(1+cosx)])
= 2sin(x)/[(1+cosx)^2]
Therefore, the derivative of y=(sinx/(1+cosx))^2 is dy/dx = 2sin(x)/[(1+cosx)^2].