Differentiate??

Question

f(x) = sin x / [ 1 + cos x ]

f(x)= 3 sin x cos x

f(x)= cos x * sin x



what the crap?? I dont know how to do it. Some one show me PLEASE!!!

Answer 1

f(x) = sin x / [ 1 + cos x ]

For this we'll use the division rule for diff.
Let's give the numerator and denomenator new names.
Call sin x by f(x) and [1+ cosx] by g(x).
Then f(x)= sin x and g(x) = [1+ cos x]
And the derivatives, f' and g' are
f'= (cos x) and g'=[0+(-sinx)] = (-sinx)
The formula is (gf' - g'f)/ g^2
so then, {[1+cosx](cosx) - (-sinx)[sinx]}/ [1+cosx]^2
Multiply the top out and try to simplify. Don't multiply out the bottom though, you'll see why later.
We get { cos x + (cosx)^2 + (sinx)^2 } / [1+cos x]^2
Notice that from trig we have that (cosx)^2 + (sinx)^2 = 1
So using this fact, we have
[cos x +1] / [1+ cos x]^2
and we can cancel out one of the bottom terms with the top. (addition is associative)
And so our final result is
1 / [1+ cos x].
We also could have used the multiplication rule using negative exponents.
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f(x)= cos x * sin x

Notice this is the same as f(x)= sin x cos x
We'll use the multiplication rule.
Let's call the functions f and g.
So let f = sin x and g = cos x then
d/dx(f*g) = gf'+g'f
and so [cosx](cosx) + (-sin x)[sinx]
Then we have
[cos x]^2 - [sinx]^2
Thus,
d/dx[sinx cos x] = [cos x]^2 - [sinx]^2
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f(x)= 3 sin x cos x

This function is the same as the previous function except for the multipler 3. From the properties of derivatives, we can take the constant, 3, out in front of the derivative.
So then,
d/dx[ 3 sinx cos x] = 3* d/dx[sinx cos x] = 3{[cos x]^2 - [sinx]^2}
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Hope that helps.

Answer 2

Question 1
using quotient rule:-
f ` (x) = (1 + cosx).cosx + sinx.sinx ) / (1 + cosx)²
f `(x) = (cosx + cos²x + sin²) / (1 + cosx)²
f ` (x) = (1 + cos x) / (1 + cos x)²
f ` (x) = 1 / (1 + cos x)

Question 2
Use product rule
f(x) = (3sinx).(cosx)
f ` (x) = 3cosx.cosx - 3.sinx.sinx
f ` (x) = 3.(cos²x - sin²x)
f ` (x) = 3.cos 2x

Question 3
Product rule:-
f(x) = cosx.sinx
f ` (x) = -sinx.sinx + cosx.cosx
f ` (x) = cos²x - sin²x
f ` (x) = cos 2x

Answer 3

1.
dy/dx =(1+cos x)(cos x) - (sin x) (-sin x) / (1+cos x) ^ 2
=cos x +cos^2 x +sin^2 x / (1+cos x) ^ 2
= cos x +1 / (1+cos x) ^ 2 (since sin ^2 x +cos^2 x=1)
= 1 / (1+cos x)

2.
dy/dx = 3sinx(-sin x) + cosx(3 cos x)
= -3sin ^2 x+3 cos^2 x
= 3(cos^2 x- sin^2x)
= 3(cos2x)
3.
dy/dx=cos x(cosx) + sin x (- sin x)
= cos^2 x-sin^2 x
= cos 2x...................................ans