derivative of (4x+3+(-8x-1))

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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 ((4x+3)+(−8x−1))

Solution

−4

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ddx ((4x+3)+(−8x−1))

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

=ddx (4x)+ddx (3)−ddx (8x)−ddx (1)

ddx (4x)=4

ddx (3)=0

ddx (8x)=8

ddx (1)=0

=4+0−8−0

Simplify

=−4

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

derivative of (4x+3+(-8x-1))

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