Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

y=ln(3x^2 +x) solve y^((6)) (x)

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

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

we have y^′ (x)=((6x+1)/(3x^2 +x)) =((6x+1)/(x(3x+1))) =(a/x) +(b/(3x+1)) a=1 and b=((6(−(1/3))+1)/((−(1/3)))) =(−3)(−1) =3 ⇒y^′ (x)=(1/x) +(3/(3x+1)) =(1/x) +(1/(x+(1/3))) ⇒y^((6)) (x)=(y^′ (x))^((5)) =((1/x) +(1/(x+(1/3))))^((5)) =((1/x))^((5)) +((1/(x +3^(−1) )))^((5)) =(((−1)^5 5!)/x^6 ) +(((−1)^5 5!)/((x+3^(−1) )^6 )) ⇒ y^((6)) (x)=−5!{(1/x^6 ) +(1/((x+3^(−1) )^6 ))} .

$${we}\:{have}\:{y}^{'} \left({x}\right)=\frac{\mathrm{6}{x}+\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} \:+{x}}\:=\frac{\mathrm{6}{x}+\mathrm{1}}{{x}\left(\mathrm{3}{x}+\mathrm{1}\right)}\:=\frac{{a}}{{x}}\:+\frac{{b}}{\mathrm{3}{x}+\mathrm{1}} \\ $$$${a}=\mathrm{1}\:\:{and}\:{b}=\frac{\mathrm{6}\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{1}}{\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)}\:=\left(−\mathrm{3}\right)\left(−\mathrm{1}\right)\:=\mathrm{3}\:\Rightarrow{y}^{'} \left({x}\right)=\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{3}}{\mathrm{3}{x}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{\mathrm{3}}}\:\Rightarrow{y}^{\left(\mathrm{6}\right)} \left({x}\right)=\left({y}^{'} \left({x}\right)\right)^{\left(\mathrm{5}\right)} =\left(\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{\mathrm{3}}}\right)^{\left(\mathrm{5}\right)} \\ $$$$=\left(\frac{\mathrm{1}}{{x}}\right)^{\left(\mathrm{5}\right)} \:+\left(\frac{\mathrm{1}}{{x}\:+\mathrm{3}^{−\mathrm{1}} }\right)^{\left(\mathrm{5}\right)} =\frac{\left(−\mathrm{1}\right)^{\mathrm{5}} \mathrm{5}!}{{x}^{\mathrm{6}} }\:+\frac{\left(−\mathrm{1}\right)^{\mathrm{5}} \mathrm{5}!}{\left({x}+\mathrm{3}^{−\mathrm{1}} \right)^{\mathrm{6}} }\:\Rightarrow \\ $$$${y}^{\left(\mathrm{6}\right)} \left({x}\right)=−\mathrm{5}!\left\{\frac{\mathrm{1}}{{x}^{\mathrm{6}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{3}^{−\mathrm{1}} \right)^{\mathrm{6}} }\right\}\:. \\ $$

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

you are welcome

$${you}\:{are}\:{welcome} \\ $$

Commented by 20190927 last updated on 25/Oct/19

thank you

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

Answered by Joel578 last updated on 24/Oct/19

y′ = ((6x + 1)/(3x^2 + x)) Let f(x) = 3x + 1, g(x) = 3x^2 + x y′′ = ((f ′g − fg′)/g^2 ) y′′′ = (((f ′g − fg′)′(g^2 ) − (f ′g − fg′)(2gg′))/g^4 ) = ((g^2 (f ′′g − fg′′) − 2gg′(f ′g − fg′))/g^4 ) I think you can continue with yourself

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

Commented by 20190927 last updated on 25/Oct/19

thank you

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com