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3*4/7

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

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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ddx ((x²−4)(5x+16))

Solution

15x²+32x−20

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ddx ((x²−4)(5x+16))

Appliquer la règle du produit: (f·g)′=f ′·g+f·g′

f=x²−4, g=5x+16

=ddx (x²−4)(5x+16)+ddx (5x+16)(x²−4)

ddx (x²−4)=2x

ddx (5x+16)=5

=2x(5x+16)+5(x²−4)

Simplifier 2x(5x+16)+5(x²−4): 15x²+32x−20

=15x²+32x−20

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

3*4/7

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