math help!
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25 May 2011, 20:48
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Journals
help me out smart cf community :D
f'(x)= ln sin (3x - 2) ???
the answer is : 3 cotg (3x-2) (but i don't get how i get there)
i know that ln f'(x) = f'(x)/f(x)
so that would mean: (sin (3x-2))' / sin (3x-2) = 3 cos (3x-2)/sin (3x-2)
or am i wrong ?
where the hell does the cotg come from since i dont have 2x sin in the denominator
f'(x)= ln sin (3x - 2) ???
the answer is : 3 cotg (3x-2) (but i don't get how i get there)
i know that ln f'(x) = f'(x)/f(x)
so that would mean: (sin (3x-2))' / sin (3x-2) = 3 cos (3x-2)/sin (3x-2)
or am i wrong ?
where the hell does the cotg come from since i dont have 2x sin in the denominator
f'(x) = d ln(sin3x-2) / dx
d ln f(x) / dx = f'(x) / f(x)
d sin(3x-2) / dx = 3cos(3x-2)
all together is 3cos(3x-2)/sin(3x-2) = 3cot(3x-2)
**please note i haven't done this maths in a while**
since cos/sin = cotg i guess thats why the answer is 3 cotg (3x-2)
i believe your solution is wrong :p since i got all solutions, i just cant find how to get there sometimes
but thx for the reply!
d f(g(x)) /dx = f'(g(x)) g'(x)
f(g(x)) = ln (sin(3x-2))
d g(x) / dx = 3cos(3x-2)
f'(x) = 1 / x
f'(g(x)) = 1 / sin(3x-2)
f'(g(x)) g'(x) = 3cos(3x-2) / sin(3x-2)
is that better?
thx for the help :)
if you state f'(x) = ln (sin (3x-2))
then what's the question?? f(x)?? f''(x)??