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intervalo (x^2+1)(x^2-1)

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derivada implícita d²ydx² , x³+y³=4

Solution

d²ydx² =−2x(x³+y³)y⁵

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x³+y³=4

Tratar y como y(x)

Derivada implícita dydx de x³+y³=4: −x²y²

dydx =−x²y²

Derivada implícita d²ydx² de dydx =−x²y² : −2x(−xdydx +y)y³

d²ydx² =−2x(−xdydx +y)y³

Substituir dydx =−x²y² : −2x(x³+y³)y⁵

d²ydx² =−2x(x³+y³)y⁵

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

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intervalo (x^2+1)(x^2-1)

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