(sin^2(theta))'

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

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(sin²(θ))^′

Solution

sin(2θ)

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(sin²(θ))′

Apply the chain rule: 2sin(θ)(sin(θ))′

=2sin(θ)(sin(θ))′

(sin(θ))′ =cos(θ)

=2sin(θ)cos(θ)

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

=sin(2θ)

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

(sin^2(theta))'

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