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limit as x approaches-infinity of (1.001)^x

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

☐²

x^☐

√☐

^□√☐

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log_□

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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 (4x²−2x+12x+3 )

Solution

8x²+24x−8(2x+3)²

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ddx (4x²−2x+12x+3 )

(fg )^′=f ′·g−g′·fg² :استعمل قانون القسمة

=ddx (4x²−2x+1)(2x+3)−ddx (2x+3)(4x²−2x+1)(2x+3)²

ddx (4x²−2x+1)=8x−2

ddx (2x+3)=2

=(8x−2)(2x+3)−2(4x²−2x+1)(2x+3)²

(8x−2)(2x+3)−2(4x²−2x+1) وسٌع: 8x²+24x−8

=8x²+24x−8(2x+3)²

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

limit as x approaches-infinity of (1.001)^x

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