Question

\[\mathrm D^6\big(x^3e^{ax}\big)\]

Answer 1

\[\begin{equation*}\mathrm D \equiv \frac{\mathrm d}{\mathrm dx}\end{equation*}
\]

Answer 2

\[\begin{align*} \mathrm D^6\big(x^3e^{ax}\big)&\\
&=\mathrm D^5\big(3x^2+ax^3\big)e^{ax}\\
&=\mathrm D^4\big(6x+3ax^2+3ax^2+a^2x^3\big)e^{ax}\\
&=\mathrm D^4\big(6x+6ax^2+a^2x^3\big)e^{ax}\\
&=\mathrm D^3\big(6+12ax+3a^2x^2+6ax+6a^2x^2+a^3x^3\big)e^{ax}\\
&=\mathrm D^3\big(6+18ax+9a^2x^2+a^3x^3\big)e^{ax}\\
&=\mathrm D^2\big(18a+18a^2x+3a^3x^2+6a+18a^2x+9a^3x^2+a^4x^3\big)e^{ax}\\
&=\mathrm D^2\big(24a+36a^2x+12a^3x^2+a^4x^3\big)e^{ax}\\
&=\mathrm D\big(36a^2+24a^3x+3a^4x^2+24a^2+36a^3x+12a^4x^2+a^5x^3\big)e^{ax}\\
&=\mathrm D\big(60a^2+60a^3x+15a^4x^2+a^5x^3\big)e^{ax}\\
&=\big(60a^3+30a^4x+3a^5x^2+60a^3+60a^4x+15a^5x^2+a^6x^3\big)e^{ax}\\
&=\big(120a^3+90a^4x+18a^5x^2+a^6x^3\big)e^{ax}\\
&=\big(120+90ax+18a^2x^2+a^3x^3\big)a^3e^{ax}\\
\end{align*}\]

Answer 3

Is there an easier way to do this ?

Answer 4

D is derivative operator ?
can't u use general formula for \(D^n(x^me^{at})\)

Answer 5

*ax

Answer 6

\[\mathrm D^n(x^me^{ax})=?\]

Answer 7

i thought there was general formula....but apparently there isn't xD

Answer 8

some thing with factorials !

Answer 9

tried Leibnitz theorem ? \(D^n(uv)\)

Answer 10

what is that?

Answer 11

https://docs.google.com/viewer?a=v&q=cache:9WaUs5ovMl8J:www.math.osu.edu/~nevai.1/H16x/DOCUMENTS/leibniz_product_formula_H6.pdf+&hl=en&gl=in&pid=bl&srcid=ADGEEShrDfDyJl8y40cSZzEHIU0yJcWw6JunzRb9a0fIS20lbReliJzWWroyHm2gc_YX4UUZ5J_bQaXUfaa1e80MB67e9UYKKMnKVVYQZmrG0G4oWkzj1wshx_2H4Pbh6HtYYhruMq2y&sig=AHIEtbTOMTfpYiFiWbXziBk9x2R7lGVEfA

Answer 12

oh

Answer 13

does that become easier ? i doubt :P

Answer 14

\[ D^6(x^3 e^{ax})=e^{ax}D^6(x^3)+6D^5(x^3)D(e^{ax}) +\cdots+6D(x^3)D^5(e^{ax})+x^3D^6(e^{ax})\]\[D^6x^3=0, \;D^5x^3=0,\; D^4x^3=0,\;D^3x^3=3!\]

Answer 15

\[D^2x^3=6x,\;Dx^3=3x^2,\;D^n(e^{ax})=a^ne^{ax}\]\[D^6(x^3e^{ax})=20(3!a^3e^{ax})+15(6xa^4e^{ax})+6(3x^2a^5e^{ax})+(x^3a^6e^{ax})\]it appears the solution is shorter

Answer 16

i guess Leibniz theorem is better.