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Question Number 72044 by 20190927 last updated on 23/Oct/19

$$\mathrm{y}=\ln ({3\mathrm{x}}^2+\mathrm{x})\mathrm{solve}{\mathrm{y}}^{\left(6\right)}\left(\mathrm{x}\right)$$

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

$$wehave{y}^{{}^{\prime }}\left(x\right)=\frac{6x+1}{3x^2+x}=\frac{6x+1}{x(3x+1)}=\frac{a}{x}+\frac{b}{3x+1}$$$$a=1andb=\frac{6(-\frac{1}{3})+1}{(-\frac{1}{3})}=(-3)(-1)=3\Rightarrow y^{{}^{\prime }}\left(x\right)=\frac{1}{x}+\frac{3}{3x+1}$$$$=\frac{1}{x}+\frac{1}{x+\frac{1}{3}}\Rightarrow y^{\left(6\right)}\left(x\right)={\left(y^{{}^{\prime }}\left(x\right)\right)}^{\left(5\right)}={(\frac{1}{x}+\frac{1}{x+\frac{1}{3}})}^{\left(5\right)}$$$$={\left(\frac{1}{x}\right)}^{\left(5\right)}+{\left(\frac{1}{x+3^{-1}}\right)}^{\left(5\right)}=\frac{{(-1)}^{5}5!}{x^6}+\frac{{(-1)}^{5}5!}{{(x+3^{-1})}^6}\Rightarrow$$$$y^{\left(6\right)}\left(x\right)=-5!\{\frac{1}{x^6}+\frac{1}{{(x+3^{-1})}^6}\}.$$

Commented by mathmax by abdo last updated on 25/Oct/19

$$youarewelcome$$

Commented by 20190927 last updated on 25/Oct/19

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

Answered by Joel578 last updated on 24/Oct/19

$$y^{\prime }=\frac{6x+1}{3x^2+x}$$$$\mathrm{Let}f\left(x\right)=3x+1,g\left(x\right)=3x^2+x$$$$$$$$y^{″}=\frac{f{}^{\prime }g-{fg}^{\prime }}{g^2}$$$$y^{‴}=\frac{{(f{}^{\prime }g-{fg}^{\prime })}^{\prime }\left(g^2\right)-(f{}^{\prime }g-{fg}^{\prime })\left(2{gg}^{\prime }\right)}{g^4}$$$$=\frac{g^2(f{}^{″}g-{fg}^{″})-2{gg}^{\prime }(f{}^{\prime }g-{fg}^{\prime })}{g^4}$$$$\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{can}\mathrm{continue}\mathrm{with}\mathrm{yourself}$$

Commented by 20190927 last updated on 25/Oct/19

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

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