et f(x) = x^4 e^(-6 x) Find a formula for the first and second derivatives of f. f'(x) = eAx (Bx4 + Cx3) and f''(x) = eDx (Ex4 + Fx3 + Gx2) where i think they want me to find A.B.C.D.E.F.G
Do you know how to find the first derivative?
\[f(x)=x^4 \cdot e^{-6x} \text{ use product rule please}\] \[f'(x)=(x^4)' \cdot e^{-6x}+x^4 \cdot (e^{-6x})' \]
I wait to hear what you think is the next step...
sorry i was trying the problem
u use the product rule to find the second derivative?
for the first derivative dont u factor out like terms?
\[f'(x)=4x^3 e^{-6x}+x^4(-6)e^{-6x}=x^3e^{-6x}(4-6x)\] \[f''(x)=(x^3)' \cdot e^{-6x}(4-6x)+x^3 \cdot (e^{-6x})' \cdot (4-6x)+x^3e^{-6x} \cdot (4-6x)'\] \[f''(x)=3x^2 \cdot e^{-6x}(4-6x)+x^3 \cdot (-6)e^{-6x} \cdot (4-6x)+x^3e^{-6x}(0-6)\] \[f''(x)=e^{-6x}(3x^2(4-6x)+x^3(-6)(4-6x)+x^3(-6))\] Now it is just a little bit of algebra to put it in the form they want to determine the coefficients.
wait so for the first derivative i got e^-6x(5x^4+4x^3)
yes i did a little renaming if you didn't notice
oh lol
so A=-6 b= 5 c= 4
That first line is f' Second line is f'' Third line is f'' fourth line is f''
\[C=4, B=-6 ,A=-6 \] hmmmm how did you get B=5? you have e^{-6x} right not e^{5x}
So your B is the only thing I don't agree with
ok so for the second derivative i got
d=-6, e=5,f=4,g=3
wait g= 4
e =3
E,F,G I see something totally different
im adding the like terms together, am i doing it correctly?
\[f ''(x)=e^{-6x}(3x^2(4-6x)+x^3(-6)(4-6x)+x^3(-6)) \] \[f''(x)=e^{-6x}(x^4(-6)(-6)+x^3(-18-24-6)+x^2(3(4)))\]
Yes put all your like terms together after factoring out e^{-6x}
Ok so e=36 , f=-48, g=12
am i right?
i agree with your answers
yeah
your'e better than my math teacher! lol
u should actually be my math professor ! lol
That is sweet! Thanks.
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