2007-02-12 01:25:00 · 4 answers · asked by C.T. 2 in Science & Mathematics ➔ Mathematics
f(x)=(3x+2)(2x^2-5x+1)
First Let u= 3x+2 and v=2x^2-5x+1
f'(x)= udv+vdu
therefore f'(x)=
(3x+2)*(4x-5)+(2x^2-5x+1)*3
= 12x^2-15x+8x-10+6x^2-15x+3
= 18x^2-22x-7
2007-02-12 01:36:35 · answer #1 · answered by The exclamation mark 6 · 3⤊ 0⤋
(uv)' = uv' + u'v d/dx(x^3(5x + a million)^2) = x^3(5)(2(5x + a million)) + 3x^2(5x + a million)^2 = 5x^3(10x + 2) + 3x^2(25x^2 + 10x + a million) = 50x^4 + 10x^3 + 75x^4 + 30x^3 + 3x^2 = 125x^4 + 40x^3 + 3x^2
2016-12-04 02:05:52 · answer #2 · answered by lemanski 4 · 0⤊ 0⤋
Here's how the product rule is set up:
f'(x)=f'(x)*g(x)+f(x)*g'(x)
f(x)=3x+2
g(x)=2x^2-5x+1
f'(x)=((3)*(2x^2-5x+1))+((3x+2)
(4x-5))
=(6x^2-15x+3)+(12x^2-7x-10)
=18x^2-22x-7
I hope this helps!
2007-02-12 03:10:13 · answer #3 · answered by Anonymous · 3⤊ 0⤋
if f = LR then df = LdR + RdL
f'(x)
=(3x+2)(4x-5)+(2x^2-5x+1)(3)
=(12x^2-7x-10)+(6x^2-15x+3)
=18x^2-22x-7
2007-02-12 01:38:07 · answer #4 · answered by tiffany twisted 3 · 3⤊ 0⤋