추가 주문하다 57y-y^2+(y+1)^2

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

☐²

x^☐

√☐

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log_□

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∫

ddx

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

simplify solve for inverse tangent line

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normal of y=(3x²−1)³, at x=13

Solution

y=−38 x−37216

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Solution steps

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Find the tangent point: (13 , −827 )

Find the slope of y=(3x²−1)³: 18x(3x²−1)²

Compute the slope at x=13 : m=83

Compute the slope of the perpendicular line: m_p=−38

Find the line with slope m=−38 and passing through (13 , −827 ): y=−38 x−37216

y=−38 x−37216

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

추가 주문하다 57y-y^2+(y+1)^2

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