f (x) = ln(1 - sin x)
2007-09-21 12:13:22 · 5 answers · asked by Niall R 1 in Science & Mathematics ➔ Mathematics
f `(x):-
- (cos x / (1 - sin x)) = cos x / (sin x - 1)
f "(x):-
[(sin x - 1)(- sin x) - (cos x)(cos x)] / (sin x - 1) ²
(- sin ² x + sin x - cos ² x) / (sin x - 1) ²
(- sin ² x + sin x - 1 + sin ² x) / (sin x - 1) ²
(sin x - 1) / (sin x - 1) ²
1 / (sin x - 1)
2007-09-25 10:50:50 · answer #1 · answered by Como 7 · 2⤊ 0⤋
The answer is 0.
Proof.
f(x) = ln(1-sin(x))
f'(x) = 1/(1-sin(x))*-cos(x)
= -cos(x)/(1-sin(x))
so using the quotient rule
f''(x) =(1-sin(x))sin(x) + cos(x)cos(x))/(1-sin(x))^2
=1 -sin^2(x)-cos^2(x)/(1-sin(x))^2
=1-(sin^2(x)+cos^2(x))/(1-sin(x))^2
Since sin^2(x) +cos^2(x) = 1
f''(x) = 1-1/(1-sin(x))^2
= 0
2007-09-25 17:12:32 · answer #2 · answered by mr_maths_man 3 · 0⤊ 0⤋
f(x) = ln(x)
f '(x) = 1/(x)* (derivative of x)
so
f ' (x) = 1/(1-sin(x)) *(cos(x)) = cos(x)/(1-sin(x))
f "(x) = quotient rule
(N' D - ND') / D^2
[(-sin(x))(1-sin(x)) - (cos(x))(cos(x))] / (1-sin(x))^2 =
[-sin(x) +sin^2(x) - cos^2(x)]/ (1-sin(x))^2
2007-09-21 19:23:50 · answer #3 · answered by intrepid_mesmer 3 · 1⤊ 0⤋
The function (x) means the opposite of x. Therefore, it basically means it's y. I learned that in 7th grade!
Good luck on the other variables! Sorry I can't help!
2007-09-26 20:54:23 · answer #4 · answered by Adreanna A 4 · 0⤊ 1⤋
quotient=vdu/dx-udv/dx all over v square-
2007-09-29 13:11:22 · answer #5 · answered by George 3 · 0⤊ 0⤋