derivative of (x+3(x-3))

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

☐²

x^☐

√☐

^□√☐

□□

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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ddx ((x+3)(x−3))

Solution

2x

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ddx ((x+3)(x−3))

Apply the Product Rule: (f·g)′=f ′·g+f·g′

f=x+3, g=x−3

=ddx (x+3)(x−3)+ddx (x−3)(x+3)

ddx (x+3)=1

ddx (x−3)=1

=1· (x−3)+1· (x+3)

Simplify 1· (x−3)+1· (x+3): 2x

=2x

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

derivative of (x+3(x-3))

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