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Question Number 5855 by sanusihammed last updated on 01/Jun/16

I f {x y}^{3} = s i n (x y)

f i n d \frac{d^{2} y}{{d x}^{2}}

P l e a s e i n e e d h e l p . t h a n k s f o r y o u r e f f o r t

Answered by 123456 last updated on 02/Jun/16

{x y}^{3} = \sin (x y)

\frac{d ({x y}^{3})}{d x} = \frac{d [\sin (x y)]}{d x}

\frac{d x}{d x} y^{3} + x \frac{d (y^{3})}{d x} = \frac{d [\sin (x y)]}{d (x y)} \cdot \frac{d (x y)}{d x}

y^{3} + x \frac{d (y^{3})}{d y} \cdot \frac{d y}{d x} = \cos (x y) \cdot (\frac{d x}{d x} y + x \frac{d y}{d x})

y^{3} + 3 {x y}^{2} \frac{d y}{d x} = \cos (x y) (y + x \frac{d y}{d x})

y^{3} + 3 {x y}^{2} \frac{d y}{d x} = \cos (x y) y + \cos (x y) x \frac{d y}{d x}

[3 {x y}^{2} - \cos (x y) x] \frac{d y}{d x} = \cos (x y) y - y^{3}

\frac{d y}{d x} = \frac{\cos (x y) y - y^{3}}{3 {x y}^{2} - \cos (x y) x}

continue

Commented by sanusihammed last updated on 02/Jun/16

I t i s t h e \frac{d^{2} y}{{d x}^{2}} t h a t i r e a l l y n e e d

Answered by Yozzii last updated on 02/Jun/16

{x y}^{3} = s i n (x y)

I m p l i c i t d i f f e r e n t i a t i o n \Rightarrow y^{3} + 3 {x y}^{2} y^{'} = ({x y}^{'} + y) c o s (x y) \Rightarrow y^{'} = \frac{y^{3} - y c o s (x y)}{x (c o s (x y) - 3 y^{2})} = \frac{y (y^{2} - c o s (x y))}{x (c o s (x y) - 3 y^{2})}

I m p l i c i t d i f f e r e n t i a t i o n a g a i n \Rightarrow 3 y^{2} y^{^{'}} + 3 {x y}^{2} y^{″} + (3 y^{2} + 6 {x y y}^{^{'}}) y^{^{'}}

= ({x y}^{″} + y^{'} + y^{'}) c o s (x y) - {({x y}^{'} + y)}^{2} s i n (x y)

3 {x y}^{2} y^{″} + 6 {y y}^{'} (y + {x y}^{'}) = {x y}^{″} c o s (x y) + 2 y^{'} c o s (x y) - {({x y}^{'} + y)}^{2} s i n (x y)

x (3 y^{2} - c o s (x y)) y^{″} = y^{'} (2 c o s (x y) - 6 y (y + {x y}^{'})) - {({x y}^{'} + y)}^{2} s i n (x y)

y^{″} = \frac{y^{'} (2 c o s (x y) - 6 y (y + {x y}^{'})) - {x y}^{3} {({x y}^{'} + y)}^{2}}{x (3 y^{2} - c o s (x y))}

L e t u = x (3 y^{2} - c o s (x y)), v = y (y^{2} - c o s (x y))

∴ y^{'} = \frac{v}{u}

y^{″} = \frac{\frac{v}{u} (2 c o s (x y) - 6 y (y + x \frac{v}{u})) - {x y}^{3} {(\frac{x v}{u} + y)}^{2}}{u}

y^{″} = \frac{v (2 u c o s (x y) - 6 y (y u + x v)) - {x y}^{3} {(x v + y u)}^{2}}{u^{3}}

y^{″} = \frac{2 v u c o s (x y) - 6 y v (y u + x v) - {x y}^{3} {(x v + y u)}^{2}}{u^{3}}

x v + y u = x y (y^{2} - c o s (x y)) + y x (3 y^{2} - c o s (x y))

x v + y u = 2 x y (2 y^{2} - c o s (x y))

y^{″} = \frac{2 y (y^{2} - c o s (x y)) (3 y^{2} - c o s (x y)) c o s (x y) - 12 y^{3} (2 y^{2} - c o s (x y)) (y^{2} - c o s (x y)) - 4 x^{2} y^{5} {(2 y^{2} - c o s (x y))}^{2}}{x^{2} {(3 y^{2} - c o s (x y))}^{3}}

Commented by sanusihammed last updated on 02/Jun/16

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