テイラー tan(5x)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

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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end point (0.25, 4), (−0.9, −2)

Solution

(1.4, 10)

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Solution steps

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End point given startpoint (x₁, y₁) and midpoint (x₂, y₂): (2x₂−x₁, 2y₂−y₁)

(x₁, y₁)=(−0.9, −2), (x₂, y₂)=(0.25, 4)

=(2· 0.25−(−0.9), 2· 4−(−2))

Simplify

=(1.4, 10)

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

テイラー tan(5x)

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