f(x)=(x-3)+(x+2)+(x+4)

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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

simplify solve for inverse tangent line

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ddx ((2x²−6x)+(−2x²+5x))

Solution

−1

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Solution steps

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ddx ((2x²−6x)+(−2x²+5x))

Apply the Sum/Difference Rule: (f±g)′=f ′±g′

=ddx (2x²)−ddx (6x)−ddx (2x²)+ddx (5x)

ddx (2x²)=4x

ddx (6x)=6

ddx (2x²)=4x

ddx (5x)=5

=4x−6−4x+5

Simplify

=−1

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foci f(x)=(2x²−6x)+(−2x²+5x)

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

f(x)=(x-3)+(x+2)+(x+4)

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