a) y=sec(3x)
b) y= x^2/2x^2+1
c) y=x^3
2006-10-25 11:24:41 · 4 answers · asked by Jennifer A 1 in Science & Mathematics ➔ Mathematics
a.3sec(3x)tan(3x)dx
b.(2x^2+1)(2x)-x^2(4x)/(2x^2+1)^2dx
=2x/(2x^2+1)^2dx
c.3x^2dx
2006-10-25 11:56:27 · answer #1 · answered by raj 7 · 0⤊ 0⤋
a) y = sec(3x) = 1/cos(3x)
Let u = cos(3x)
If u = cos(3x) then du = -3sin(3x)
Therefore by the quotient rule
dy = -(-3sin(3x)) / (cos(3x))^2
= 3sin(3x) / (cos(3x))^2
= 3(sin(3x)/cos(3x))(1/cos(3x))
=3 tan(3x) sec(3x)
b) This problem is ambiguous without parenthesis. Need to know if the '1' is part of the denominator or a separate expression.
c) y = x^3
dy = 3x^2
2006-10-25 18:53:16 · answer #2 · answered by Maj H 1 · 0⤊ 0⤋
a) (3sec3x*tan3x) dx
b) (2x/(2x^2+1) - (4x^3)/(2x^2+1)^2) dx
c) 3x^2 dx
2006-10-25 18:36:22 · answer #3 · answered by Steve 7 · 0⤊ 0⤋
a) 3(sec(3x)tan(3x))dx
b) (2x / ((2x^2 + 1)^2))dx
c) (3x^2)dx
2006-10-25 19:03:24 · answer #4 · answered by d@mn 2 · 0⤊ 0⤋