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


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

$y = (a^{2} + x^{2}) \tan^{- 1} \frac{x}{a}$ $F i n d \frac{d^{3} y}{{d x}^{3}} .$
Commented by math khazana by abdo last updated on 19/Jul/18

$w e h a v e y (x) = (x^{2} + a^{2}) a r c t a n (\frac{x}{a}) \Rightarrow$ $y^{(1)} (x) = 2 x a r c t a n (\frac{x}{a}) + (x^{2} + a^{2}) \frac{1}{a (\frac{x^{2}}{a^{2}} + 1)}$ $= 2 x a r c t a n (\frac{x}{a}) + (x^{2} + a^{2}) \frac{a^{2}}{a (x^{2} + a^{2})}$ $= 2 x a r c t a n (\frac{x}{a}) + a \Rightarrow$ $y^{(2)} (x) = 2 a r c t a n (\frac{x}{a}) + 2 x \frac{1}{a (\frac{x^{2}}{a^{2}} + 1)}$ $= 2 a r c t a n (\frac{x}{a}) + \frac{2 a x}{x^{2} + a^{2}} \Rightarrow$ $y^{(3)} (x) = \frac{2}{a (\frac{x^{2}}{a^{2}} + 1)} + \frac{2 a (x^{2} + a^{2}) - 2 a x (2 x)}{{(x^{2} + a^{2})}^{2}}$ $= \frac{2 a}{x^{2} + a^{2}} + \frac{2 {a x}^{2} + 2 a^{3} - 4 {a x}^{2}}{{(x^{2} + a^{2})}^{2}}$ $= \frac{2 a (x^{2} + a^{2}) - 2 {a x}^{2} + 2 a^{3}}{{(x^{2} + a^{2})}^{2}} = \frac{4 a^{3}}{{(x^{2} + a^{2})}^{2}} .$
Commented by ajfour last updated on 19/Jul/18

$T h a n k y o u S i r!$
Commented by math khazana by abdo last updated on 20/Jul/18

$n e v e r m i n d s i r .$
	Answered by MJS last updated on 19/Jul/18

	$\frac{d}{d θ} [\arctan θ] = \frac{1}{1 + θ^{2}}$ $\frac{d y}{d x} [(a^{2} + x^{2}) \arctan \frac{x}{a}] = a + 2 x \arctan \frac{x}{a}$ $\frac{d^{2} y}{{d x}^{2}} [(a^{2} + x^{2}) \arctan \frac{x}{a}] = \frac{2 a x}{a^{2} + x^{2}} + 2 \arctan \frac{x}{a}$ $\frac{d^{3} y}{{d x}^{3}} [(a^{2} + x^{2}) \arctan \frac{x}{a}] = \frac{4 a^{3}}{{(a^{2} + x^{2})}^{2}}$
	Commented by ajfour last updated on 19/Jul/18

	$G r e a t!$
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