3xy-x-2-y-2-5-find-the-second-derivative-

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} + 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 \dots

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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