d/(du)(sin^2(u))

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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simplify solve for inverse tangent line

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ddu (sin²(u))

Solution

sin(2u)

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Solution steps

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ddu (sin²(u))

Apply the chain rule: 2sin(u)ddu (sin(u))

=2sin(u)ddu (sin(u))

ddu (sin(u))=cos(u)

=2sin(u)cos(u)

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

=sin(2u)

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

d/(du)(sin^2(u))

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