d/(db)((3b)/(9b^3))

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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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ddb (3b9b³ )

Solution

−23b³

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ddb (3b9b³ )

Take the constant out: (a·f)′=a·f ′

=39 ddb (bb³ )

Apply the Quotient Rule: (fg )^′=f ′·g−g′·fg²

=39 · dbdb b³−ddb (b³)b(b³)²

dbdb =1

ddb (b³)=3b²

=39 · 1· b³−3b²b(b³)²

Simplify 39 · 1· b³−3b²b(b³)² : −23b³

=−23b³

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

d/(db)((3b)/(9b^3))

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