3xy+x^2 +y^2 =5 find the second derivative


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Question Number 74723 by necxxx last updated on 29/Nov/19

$3 x y + x^{2} + y^{2} = 5$ $f i n d t h e s e c o n d d e r i v a t i v e$
Commented by mathmax by abdo last updated on 29/Nov/19
$y^2 +3xy+x^2 −5=0 Δ=(3x)^2 −4(x^2 −5) =9x^2 −4x^2 +20 =5x^2 +20 ⇒ y_1 (x)=((−3x+(√(5x^2 +20)))/2) and y_2 (x)=((−3x−(√(5x^2 +20)))/2) y=y_1 ⇒y^′ (x)=−(3/2)+(1/2)×((10x)/(2(√(5x^2 +20)))) =−(3/2) +(5/2)(x/(√(5x^2 +20))) and y^((2)) (x)=(5/2){x (5x^2 +20)^(−(1/2)) }^((1)) =(5/2){(5x^2 +20)^(−(1/2)) +x ×(−(1/2)(10x)(5x^2 +29)^(−(3/2)) } =(5/(2(√(5x^2 +20))))−((5x)/((5x^2 +20)(√(5x^2 +20)))) y=y_2 we follow the same method...$
$y^{2} + 3 x y + x^{2} - 5 = 0$ $Δ = {(3 x)}^{2} - 4 (x^{2} - 5) = 9 x^{2} - 4 x^{2} + 20 = 5 x^{2} + 20 \Rightarrow$ $y_{1} (x) = \frac{- 3 x + \sqrt{5 x^{2} + 20}}{2} a n d y_{2} (x) = \frac{- 3 x - \sqrt{5 x^{2} + 20}}{2}$ $y = y_{1} \Rightarrow y^{^{'}} (x) = - \frac{3}{2} + \frac{1}{2} \times \frac{10 x}{2 \sqrt{5 x^{2} + 20}} = - \frac{3}{2} + \frac{5}{2} \frac{x}{\sqrt{5 x^{2} + 20}}$ $a n d y^{(2)} (x) = \frac{5}{2} {x {(5 x^{2} + 20)}^{- \frac{1}{2}}}^{(1)}$ $= \frac{5}{2} {{(5 x^{2} + 20)}^{- \frac{1}{2}} + x \times (- \frac{1}{2} (10 x) {(5 x^{2} + 29)}^{- \frac{3}{2}}}$ $= \frac{5}{2 \sqrt{5 x^{2} + 20}} - \frac{5 x}{(5 x^{2} + 20) \sqrt{5 x^{2} + 20}}$ $y = y_{2} w e f o l l o w t h e s a m e m e t h o d . . .$
	Answered by Tanmay chaudhury last updated on 29/Nov/19

	$3 x \frac{d y}{d x} + 3 y + 2 x + 2 y . \frac{d y}{d x} = 0$ $[\frac{d y}{d x} = \frac{- (3 y + 2 x)}{3 x + 2 y}$ $3 x \times \frac{d^{2} y}{{d x}^{2}} + 3 \frac{d y}{d x} + 3 \frac{d y}{d x} + 2 + 2 y . \frac{d^{2} y}{{d x}^{2}} + 2 {(\frac{d y}{d x})}^{2} = 0$ $p l s p u t v a l u e o f \frac{d y}{d x} a n d c a l c u l a t e$
	Commented by necxxx last updated on 29/Nov/19

	$T h a n k y o u s o m u c h s i r .$
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