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tangent of x^2+2xy-y^2+x=2, (1,2)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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tangent of x²+2xy−y²+x=2, (1, 2)

Solution

y=72 x−32

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Find the slope of x²+2xy−y²+x=2: −2x−1−2y2(x−y)

Compute the slope at (1, 2): m=72

Find the line with slope m=72 and passing through (1, 2): y=72 x−32

y=72 x−32

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

tangent of x^2+2xy-y^2+x=2, (1,2)

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