amplitud 2sin^5(x)

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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

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amplitude 6cos(x)

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ddx (sin²(x))

Apply the chain rule: 2sin(x)ddx (sin(x))

=2sin(x)ddx (sin(x))

ddx (sin(x))=cos(x)

=2sin(x)cos(x)

Use the Double Angle identity: 2sin(x)cos(x)=sin(2x)

=sin(2x)

Solve instead

domain 6cos(x)

range 6cos(x)

periodicity 6cos(x)

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

amplitud 2sin^5(x)

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