2006-11-15 15:46:06 · 5 answers · asked by iheartcilantro 1 in Science & Mathematics ➔ Mathematics
Derivative of [six(3x)]^x
2006-11-15 15:59:48 · update #1
y = ([sin(3x)]^x)
lny = x ln ([sin(3x)])
(1/y) dy/dx = d/dx(x ln ([sin(3x)]) )
or, dy/dx = y [ln ([sin(3x)]) ) + x d/dx(ln (sin(3x)) ) ]
now x d/dx(ln (sin(3x)) )
= x (1/ sin(3x) ) d/dx(sin(3x))
= x (1/ sin(3x) ) cos(3x)
= x cot (3x)
so, dy/dx = y[ ln(sin(3x)) + x cot (3x) )
or, dy/dx = ([sin(3x)]^x)[ ln(sin(3x)) + x cot (3x) )
2006-11-16 00:35:16 · answer #1 · answered by The Potter Boy 3 · 0⤊ 0⤋
lnsin3x + 3xcos3x / sin3x
or
lnsin3x + 3xcot3x
2006-11-16 00:11:17 · answer #2 · answered by waleed 1 · 0⤊ 0⤋
just [sin (2x + x)]^x
[sin 2x cos x + cos 2x sin x]^2
[2sinxcos^2x + cos^2xsinx - sin^2x]^x
[2sinxcos^2x + cos^2xsinx - 1 + cos^2x]^x
[cos^2x(2sinx + sinx + 1) -1]^x
2006-11-16 00:09:49 · answer #3 · answered by Anonymous · 0⤊ 0⤋
without knowing what x is, you can only represent it as it is in your question.
2006-11-15 23:54:24 · answer #4 · answered by Mr. Engineer 2 · 0⤊ 0⤋
make sin then ^x!
2006-11-15 23:50:12 · answer #5 · answered by Vladimir S 2 · 0⤊ 0⤋