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Question Number 40319 by ajfour last updated on 19/Jul/18

y=(a^2 +x^2 )tan^(−1) (x/a) Find (d^3 y/dx^3 ) .

$${y}=\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{{a}} \\ $$$${Find}\:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:. \\ $$

Commented by math khazana by abdo last updated on 19/Jul/18

we have y(x)=(x^2 +a^2 )arctan((x/a))⇒ y^((1)) (x)=2x arctan((x/a)) +(x^2 +a^2 ) (1/(a( (x^2 /a^2 ) +1))) =2x arctan((x/a)) +(x^2 +a^2 ) (a^2 /(a(x^2 +a^2 ))) =2x arctan((x/a)) +a ⇒ y^((2)) (x)= 2 arctan((x/a)) +2x (1/(a((x^2 /a^2 )+1))) =2 arctan((x/a)) +((2ax)/(x^2 +a^2 )) ⇒ y^((3)) (x)= (2/(a( (x^2 /a^2 ) +1))) +((2a(x^2 +a^2 )−2ax(2x))/((x^2 +a^2 )^2 )) = ((2a)/(x^2 +a^2 )) +((2ax^2 +2a^3 −4ax^2 )/((x^2 +a^2 )^2 )) =((2a(x^(2 ) +a^2 )−2ax^2 +2a^3 )/((x^2 +a^2 )^2 )) = ((4a^3 )/((x^2 +a^2 )^2 )) .

$${we}\:{have}\:{y}\left({x}\right)=\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right){arctan}\left(\frac{{x}}{{a}}\right)\Rightarrow \\ $$$${y}^{\left(\mathrm{1}\right)} \left({x}\right)=\mathrm{2}{x}\:{arctan}\left(\frac{{x}}{{a}}\right)\:+\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\:\:\frac{\mathrm{1}}{{a}\left(\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\mathrm{1}\right)} \\ $$$$=\mathrm{2}{x}\:{arctan}\left(\frac{{x}}{{a}}\right)\:+\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\:\frac{{a}^{\mathrm{2}} }{{a}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)} \\ $$$$=\mathrm{2}{x}\:{arctan}\left(\frac{{x}}{{a}}\right)\:+{a}\:\Rightarrow \\ $$$${y}^{\left(\mathrm{2}\right)} \left({x}\right)=\:\mathrm{2}\:{arctan}\left(\frac{{x}}{{a}}\right)\:+\mathrm{2}{x}\:\:\frac{\mathrm{1}}{{a}\left(\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\:{arctan}\left(\frac{{x}}{{a}}\right)\:+\frac{\mathrm{2}{ax}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:\Rightarrow \\ $$$${y}^{\left(\mathrm{3}\right)} \left({x}\right)=\:\frac{\mathrm{2}}{{a}\left(\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\mathrm{1}\right)}\:+\frac{\mathrm{2}{a}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)−\mathrm{2}{ax}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{2}{a}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:+\frac{\mathrm{2}{ax}^{\mathrm{2}} \:+\mathrm{2}{a}^{\mathrm{3}} \:−\mathrm{4}{ax}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{a}\left({x}^{\mathrm{2}\:} +{a}^{\mathrm{2}} \right)−\mathrm{2}{ax}^{\mathrm{2}} \:+\mathrm{2}{a}^{\mathrm{3}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:\frac{\mathrm{4}{a}^{\mathrm{3}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:. \\ $$$$ \\ $$

Commented by ajfour last updated on 19/Jul/18

Thank you Sir!

$${Thank}\:{you}\:{Sir}! \\ $$

Commented by math khazana by abdo last updated on 20/Jul/18

nevermind sir.

$${nevermind}\:{sir}. \\ $$

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 ))

$$\frac{{d}}{{d}\theta}\left[\mathrm{arctan}\:\theta\right]=\frac{\mathrm{1}}{\mathrm{1}+\theta^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}\left[\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\mathrm{arctan}\:\frac{{x}}{{a}}\right]={a}+\mathrm{2}{x}\mathrm{arctan}\:\frac{{x}}{{a}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\left[\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\mathrm{arctan}\:\frac{{x}}{{a}}\right]=\frac{\mathrm{2}{ax}}{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }+\mathrm{2arctan}\:\frac{{x}}{{a}} \\ $$$$\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\left[\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\mathrm{arctan}\:\frac{{x}}{{a}}\right]=\frac{\mathrm{4}{a}^{\mathrm{3}} }{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$

Commented by ajfour last updated on 19/Jul/18

Great!

$$\mathbb{G}{reat}! \\ $$

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