Question Number 40319 by ajfour last updated on 19/Jul/18
$${y}=\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{{a}} \\ $$$${Find}\:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:. \\ $$
Commented by math khazana by abdo last updated on 19/Jul/18
$${we}\:{have}\:{y}\left({x}\right)=\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right){arctan}\left(\frac{{x}}{{a}}\right)\Rightarrow \\ $$$${y}^{\left(\mathrm{1}\right)} \left({x}\right)=\mathrm{2}{x}\:{arctan}\left(\frac{{x}}{{a}}\right)\:+\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\:\:\frac{\mathrm{1}}{{a}\left(\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\mathrm{1}\right)} \\ $$$$=\mathrm{2}{x}\:{arctan}\left(\frac{{x}}{{a}}\right)\:+\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\:\frac{{a}^{\mathrm{2}} }{{a}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)} \\ $$$$=\mathrm{2}{x}\:{arctan}\left(\frac{{x}}{{a}}\right)\:+{a}\:\Rightarrow \\ $$$${y}^{\left(\mathrm{2}\right)} \left({x}\right)=\:\mathrm{2}\:{arctan}\left(\frac{{x}}{{a}}\right)\:+\mathrm{2}{x}\:\:\frac{\mathrm{1}}{{a}\left(\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\:{arctan}\left(\frac{{x}}{{a}}\right)\:+\frac{\mathrm{2}{ax}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:\Rightarrow \\ $$$${y}^{\left(\mathrm{3}\right)} \left({x}\right)=\:\frac{\mathrm{2}}{{a}\left(\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\mathrm{1}\right)}\:+\frac{\mathrm{2}{a}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)−\mathrm{2}{ax}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{2}{a}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:+\frac{\mathrm{2}{ax}^{\mathrm{2}} \:+\mathrm{2}{a}^{\mathrm{3}} \:−\mathrm{4}{ax}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{a}\left({x}^{\mathrm{2}\:} +{a}^{\mathrm{2}} \right)−\mathrm{2}{ax}^{\mathrm{2}} \:+\mathrm{2}{a}^{\mathrm{3}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:\frac{\mathrm{4}{a}^{\mathrm{3}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:. \\ $$$$ \\ $$
Commented by ajfour last updated on 19/Jul/18
$${Thank}\:{you}\:{Sir}! \\ $$
Commented by math khazana by abdo last updated on 20/Jul/18
$${nevermind}\:{sir}. \\ $$
Answered by MJS last updated on 19/Jul/18
![(d/dθ)[arctan θ]=(1/(1+θ^2 )) (dy/dx)[(a^2 +x^2 )arctan (x/a)]=a+2xarctan (x/a) (d^2 y/dx^2 )[(a^2 +x^2 )arctan (x/a)]=((2ax)/(a^2 +x^2 ))+2arctan (x/a) (d^3 y/dx^3 )[(a^2 +x^2 )arctan (x/a)]=((4a^3 )/((a^2 +x^2 )^2 ))](https://www.tinkutara.com/question/Q40320.png)
$$\frac{{d}}{{d}\theta}\left[\mathrm{arctan}\:\theta\right]=\frac{\mathrm{1}}{\mathrm{1}+\theta^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}\left[\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\mathrm{arctan}\:\frac{{x}}{{a}}\right]={a}+\mathrm{2}{x}\mathrm{arctan}\:\frac{{x}}{{a}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\left[\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\mathrm{arctan}\:\frac{{x}}{{a}}\right]=\frac{\mathrm{2}{ax}}{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }+\mathrm{2arctan}\:\frac{{x}}{{a}} \\ $$$$\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\left[\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\mathrm{arctan}\:\frac{{x}}{{a}}\right]=\frac{\mathrm{4}{a}^{\mathrm{3}} }{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$
Commented by ajfour last updated on 19/Jul/18
$$\mathbb{G}{reat}! \\ $$