Use logarithmic differenctiation to compute the derivative, given y = (2x+1)^3(3x+4)^5
2006-10-27 16:18:48 · 5 answers · asked by Nina Karina 1 in Science & Mathematics ➔ Mathematics
y = (2x+1)^3(3x+4)^5
y' = 15(2x+1)^3(3x+4)^4 + 6(2x+1)^2(3x+4)^5
y' = 3(2x+1)^2(3x+4)^4 (5(2x+1) + 2(3x+4) )
2006-10-27 16:49:01 · answer #1 · answered by M. Abuhelwa 5 · 0⤊ 0⤋
ln y = ln ( 2x + 1 )^3 ( 3x+4)^5
ln y = ln ( 2x + 1)^3 + ln ( 3x+4 )^3
ln y = 3 ln ( 2x+1 ) + 5 ln ( 3x+4 )
y' (ln y) = 3 * ( 2 / 2x + 1 ) + 5 * ( 3 / 3x + 4 )
y' (ln y) = 3 *[ ( 2 / 2x+1 ) + ( 5 / 3x+4 ) ]
y' = 3 ( 2x+1 )^3 ( 3x+4 )^5 [ ( 2 / 2x+a) + ( 5 / 3x+4 ) ]
2006-10-27 23:34:08 · answer #2 · answered by c00kies 5 · 0⤊ 0⤋
ln y = ln [(2x+1)^2(3x+4)^5]
ln y = 2 ln(2x+1) + 5 ln(3x+4)
y' / y = 2*2/(2x+1) + 3*5/(3x+4)
y' = [(2x+1)^3(3x+4)^5] [ 4/(2x+1) + 15/(3x+4) ]
2006-10-27 23:28:14 · answer #3 · answered by James L 5 · 0⤊ 0⤋
i would help you but i'm just learning that too.
2006-10-27 23:58:18 · answer #4 · answered by jake 5 · 0⤊ 0⤋
make along..
2006-10-27 23:22:14 · answer #5 · answered by Morocha 6 · 0⤊ 0⤋