... calculus (1)... if y=((1/(sin(x)+cos(x))))^(1/3) then : 3y′′−12(y′)^2 −y^2 = ??? .........


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Question Number 132104 by mnjuly1970 last updated on 11/Feb/21

$. . . c a l c u l u s (1) . . .$ $i f y = \sqrt[3]{\frac{1}{s i n (x) + c o s (x)}} t h e n :$ $3 y^{″} - 12 {(y^{'})}^{2} - y^{2} = ? ? ?$ $. . . . . . . . .$
	Answered by mnjuly1970 last updated on 11/Feb/21

	$s o l u t i o n ::$ $\frac{1}{y^{3}} = s i n (x) + c o s (x)$ $d i f f e r e n t i a t i o n b o t h s i d e s : \frac{- 3 y^{'} y^{2}}{y^{6}} = c o s (x) - s i n (x)$ $s i m p l i f i c a t i o n :: \frac{3 y^{'}}{y^{4}} = s i n (x) - c o s (x)$ $d i f f e r e n t i a t i o n a g a i n :: \frac{3 y^{″} y^{4} - 12 {(y^{'})}^{2} y^{3}}{y^{8}} = c o s (x) + s i n (x) = \frac{1}{y^{3}}$ $s i m p l i f i c a t i o n :: \frac{3 y^{″} y - 12 {(y^{'})}^{2}}{y^{5}} = \frac{1}{y^{3}}$ $∴ 3 y^{″} y - 12 {(y^{'})}^{2} - y^{2} = 0$
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