If xy^(3 ) = sin(xy) find (d^2 y/dx^2 ) Please i need help. thanks for your effort


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
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

	$T h a n k s s o m u c h$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum