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

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ddx (5x+27x 2x )

Solution

−5x−414x³

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ddx (5x+27x 2x )

Entferne die Konstante: (a·f)′=a·f ′

=17 2 ddx (5x+2x x )

Wende die Quotientenregel an: (fg )^′=f ′·g−g′·fg²

=17 2 · ddx (5x+2x )x−dxdx 5x+2x x²

ddx (5x+2x )=−2x²

dxdx =1

=17 2 · (−2x² )x−1· 5x+2x x²

Vereinfache 17 2 · (−2x² )x−1· 5x+2x x² : −5x−414x³

=−5x−414x³

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

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