derivative?

Question

find the derivative of

y = ((sin x)^2)/((1+cos x)^2)

Answer 1

((1 + cosx))*(2sinxcosx) - ((sinx)^2)*(2 + 2cosx)sinx
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(1 + cosx)^4

Answer 2

I would use the quotient rule and chain rule for this one, but the work would be beyond exaggeration.

Quotient rule:

Derivative of f(x)/g(x) = [ f'(x)g(x) - g'(x)f(x) ] / g(x)^2

d/dx(sin x)^2 = 2(sin x)(cos x)

d/dx(1+cos x)^2 = 2(1+cos x)(-sin x)

Simplify 2(1+cos x)(-sin x)

-2sin x - 2cos x * sin x

Despite all that, we also have to multiply.

2(sin x)(cos x) * (1+cos x)^2 - (-2sin x - 2cos x * sin x)*(sin x)^2 / (1+cos x^4)

Whatever it is, just find it yourself...

Pretty simple to just plug in values.

Just for your information: Don't try finding the answer on computer, use paper, because computers use up more space and the work is more atrocious.

To Kundoggy: The formula you just gave for quotient rule is flipped backwards (sort of).

Answer 3

the formula here is fg'-gf'/g^2

{ (sinx)^2)* 2(1+cosx) (-sin x) - (1+cos x)^2 * 2(sin x) (cos x) } /
((1+cos x) ^2)^2

you have to use the chain rule when doing the dervitive of sinx ^2 and 1+ cos X ^2...which i did....good luck hope it helped

Answer 4

f(x) = sin² x / [ (1 + cos x) (1 + cos x) ]
= (1 - cos ² x) / [ (1 + cos x) (1 + cos x) ]
= (1 - cos x) (1 + cos x) / (1 + cos x) (1 + cos x)
= (1 - cos x) / (1 + cos x)
Using quotient rule:-
f `(x) =
(1 + cos x)(sin x) + (1 - cos x)(sinx) / (1 + cosx)²
= sin x [ 1 + cos x + 1 - cos x ] / (1 + cos x)²
= 2 sin x / (1 + cos x)²