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area (x-4)^2+(y+2)^2=25

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

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extreme points 3x²y+y³−3x²−3y²+2

Solution

Minimum(0, 2), Maximum(0, 0), Saddle(1, 1), Saddle(−1, 1)

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Solution steps

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∂ ∂ x (ln(x²+y²))

Treat y as a constant

Apply the chain rule: 1x²+y² ∂ ∂ x (x²+y²)

=1x²+y² ∂ ∂ x (x²+y²)

∂ ∂ x (x²+y²)=2x

=1x²+y² · 2x

Simplify 1x²+y² · 2x: 2xx²+y²

=2xx²+y²

Solve instead

critical points 3x²y+y³−3x²−3y²+2

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

area (x-4)^2+(y+2)^2=25

Solution

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