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Tue 9 Jul, 2013 06:45 am
Differentiate (x^4-3 x) e^x
@charsha,
Is this a nodular plan, or are we following the binary cause?
@charsha,
(f*g)' = f'g + fg'
f = x^4 - 3x
g = e^x
f' = 4x^3 -3
g' = e^x
f'g + fg' = (4x^3 - 3)e^x + (x^4-3x)e^x
You can combine those terms if you want.