2007-10-08 04:52:12 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I'll bring up the (3+7x) and use the product rule.
(e^x)*(3+7x)^-1
y'(f*g) = f'g + g'f
y' = (e^x)*(3+7x)^-1 -7*(3+7x)^-2*(e^x)
=(e^x)/(3x+7x) - (7e^x/(3+7x)^2)
2007-10-08 04:58:44 · answer #1 · answered by de4th 4 · 1⤊ 1⤋
Differentiate then simplify by combining fractions!
dy/dx = (e^x) * (7x-4) / [(3+7x)²]
These other people missed a simple simplification.
2007-10-08 11:58:07 · answer #2 · answered by 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10 6 · 0⤊ 1⤋
dy/dx
= [ (e^x) (3 + 7x) - (e^x) (7) ] / ( 3 + 7x ) ²
= [ e^x (3 + 7x - 7)] / (3 + 7x) ²
= [ e^x (7x - 4) ] / (3 + 7 x) ²
2007-10-08 12:16:57 · answer #3 · answered by Como 7 · 1⤊ 0⤋
y = (e^x) * (3+7x)^(-1)
thus
y' = {(e^x) * (-1) * [(3+7x)^(-2)] * 7 }+ (3+7x)^(-1) * (e^x)
= { (-7) (e^x) / [(3+7x)^2] } - { (e^x) / (3+7x) }
2007-10-08 11:57:23 · answer #4 · answered by sunny 4 · 0⤊ 2⤋
dy/dx = [(3 + 7x)(e^x) - 7e^x]/(3 + 7x)^2
2007-10-08 11:56:55 · answer #5 · answered by 4GET IT!!!! 2 · 0⤊ 2⤋
DO YOUR WORK YOUSELF. DONT ASK THIS TO OTHERS.
2007-10-08 12:05:09 · answer #6 · answered by VINEET S 2 · 1⤊ 0⤋