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Question Number 76126 by Master last updated on 24/Dec/19

Commented by benjo last updated on 24/Dec/19

Commented by benjo last updated on 24/Dec/19

$$\mathrm{i}\mathrm{cannot}\mathrm{find}\mathrm{formula}\mathrm{the}\mathrm{simple}\mathrm{form}\mathrm{n}-\mathrm{th}$$$$\mathrm{derivative}$$

Commented by Master last updated on 24/Dec/19

$$\mathrm{Mr}.\mathrm{n}\mathrm{degree}\mathrm{output}\mathrm{is}\mathrm{required}.\mathrm{thank}\mathrm{you}$$

Commented by MJS last updated on 24/Dec/19

$$\frac{d}{dx}[y={\mathrm{e}}^{\alpha x}({ax}^4+{bx}^3+{cx}^2+dx+e)]=$$$$={\mathrm{e}}^{\alpha x}(a\alpha x^4+(4a+b\alpha )x^3+(3b+c\alpha )x^2+(2c+d\alpha )x+d+e\alpha )$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{to}\mathrm{find}\mathrm{formulas}\mathrm{for}\mathrm{the}\mathrm{constant}$$$$\mathrm{factors},\mathrm{sorry}\mathrm{I}\mathrm{have}\mathrm{no}\mathrm{time}\mathrm{right}\mathrm{now}$$

Commented by john santuy last updated on 24/Dec/19

$$let{x}^4+x^3+1=uthen$$$$y^{{}^{\prime }}=(-2u+u^{{}^{\prime }})e^{-2x}$$$$y^{″}=(4u-2u^{{}^{\prime }})e^{-2x}+(-2u^{{}^{\prime }}+u^{{}^{″}})e^{-2x}$$$$y^{{}^{″}}=(u^{{}^{″}}-4u^{{}^{\prime }}+4u)e^{-2x}$$$$y^{{}^{‴}}=(-2u^{{}^{″}}+8u^{{}^{\prime }}-8u)e^{-2x}+(u^{{}^{‴}}-4u^{{}^{″}}+4u^{{}^{\prime }})e^{-2x}$$

Commented by john santuy last updated on 24/Dec/19

$$maybesircantheformulasn-thderivative$$

Commented by mathmax by abdo last updated on 24/Dec/19

$$letf\left(x\right)=e^{-2x}andg\left(x\right)=x^4+x^3+1\Rightarrow$$$$y\left(x\right)=f\left(x\right)g\left(x\right)\Rightarrow y^{\left(n\right)}\left(x\right)={\left(f\left(x\right)g\left(x\right)\right)}^{\left(n\right)}$$$$=\sum _{k=0}^n{C}_n^k{g}^{\left(k\right)}\left(x\right)f^{(n-k)}\left(x\right)$$$$=C_n^{0}g\left(x\right)f^{\left(n\right)}\left(x\right)+C_n^1{g}^{\left(1\right)}\left(x\right)f^{(n-1)}\left(x\right)+C_n^2{g}^{\left(2\right)}\left(x\right)f^{(n-2)}\left(x\right)+C_n^3{g}^{\left(3\right)}\left(x\right)f^{(n-3)}\left(x\right)$$$$C_n^4{g}^{\left(4\right)}\left(x\right)f^{(n-4)}\left(x\right)+...+g^{\left(n\right)}\left(x\right)f\left(x\right)$$$$={(-2)}^n(x^4+x^3+1)e^{-2x}+C_n^1(4x^3+3x^2){(-2)}^{n-1}{e}^{-2x}+$$$$+C_n^2(12x^2+6x){(-2)}^{n-2}{e}^{-2x}+C_n^3(24x+6){(-2)}^{n-3}{e}^{-2x}$$$$+C_n^{4}24{(-2)}^{n-4}{e}^{-2x}$$

Commented by Master last updated on 24/Dec/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by turbo msup by abdo last updated on 24/Dec/19

$$youarewelcome.$$

Answered by mr W last updated on 24/Dec/19

$$y=e^{-2x}(x^4+x^3+1)=u\left(x\right)v\left(x\right)$$$$acc.to{Leibnitz}^{\prime }stheorem$$$$y^{\left(n\right)}=\underset{k=0}{\sum ^n}{C}_k^n{u}^{(n-k)}{v}^{\left(k\right)}$$$$y^{\left(n\right)}={(-2)}^n{e}^{-2x}(x^4+x^3+1)$$$$+n{(-2)}^{n-1}{e}^{-2x}(4x^3+3x^2)$$$$+\frac{n(n-1)}{2}{(-2)}^{n-2}{e}^{-2x}(12x^2+6x)$$$$+\frac{n(n-1)(n-2)}{6}{(-2)}^{n-3}{e}^{-2x}(24x+6)$$$$+\frac{n(n-1)(n-2)(n-3)}{24}{(-2)}^{n-4}{e}^{-2x}\left(24\right)$$

Commented by Master last updated on 24/Dec/19

$$\mathrm{thanks}\mathrm{sir}$$

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