Question

Find the derrivative of (3-4e^x)/x^3

So far I have gotten ((-4e^x) (x^3)-3(3x^2))/x^6 is this correct?

Answer 1

no

Answer 2

close but the second term in the numerator is wrong

Answer 3

\[\left(\frac{f}{g}\right)'=\frac{f'g-g'f}{g^2}\]

Answer 4

\(f(x)=3-4e^x,f'(x)=-4e^x,g(x)=x^3,g'(x)=3x^2\)

Answer 5

unless you did some algebra maybe lets see

Answer 6

\[\frac{x^3\times (-4e^x)-3x^2(3-4e^x)}{x^6}\] is the first step

Answer 7

i was wrong,
cancel a common factor of \(x^2\) and get

\[\frac{x\times (-4e^x)-3(3-4e^x)}{x^4}\]

Answer 8

then multiply out in the numerator

Answer 9

Ohh..I think I tried taking out the -4e^x on the numerator..

Answer 10

So would the answer be x(-4e^x)-9+12e^x?

Answer 11

that looks beter

Answer 12

for the numerator

Answer 13

Alright thanks a lot you were a lot of help! :) Annd yess it'd be all over x^4 :) I almost forgot that part ha

Answer 14

yw