inverse of 2/x

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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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ddx ((x+1)(x²−2x+4))

Solution

3x²−2x+2

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Solution steps

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ddx ((x+1)(x²−2x+4))

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

f=x+1, g=x²−2x+4

=ddx (x+1)(x²−2x+4)+ddx (x²−2x+4)(x+1)

ddx (x+1)=1

ddx (x²−2x+4)=2x−2

=1· (x²−2x+4)+(2x−2)(x+1)

Simplify 1· (x²−2x+4)+(2x−2)(x+1): 3x²−2x+2

=3x²−2x+2

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

inverse of 2/x

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