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nghịch đảo 1/(x^3)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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đạo hàm ẩn d²ydx² , x³+y³=4

Solution

d²ydx² =−2x(x³+y³)y⁵

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x³+y³=4

Coi y như y(x)

Phép lấy vi phân tìm đạo hàmdydx của x³+y³=4: −x²y²

dydx =−x²y²

Phép lấy vi phân tìm đạo hàmd²ydx² của dydx =−x²y² : −2x(−xdydx +y)y³

d²ydx² =−2x(−xdydx +y)y³

Thay 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`

nghịch đảo 1/(x^3)

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