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inverse of 20((x-3)(x+4))/((x-3)^2(x-5))

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

☐²

x^☐

√☐

^□√☐

□□

log_□

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∫

ddx

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

implicit derivative chain product quotient sum/diff

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ddx (2x+xsin(x))

Solution

2+sin(x)+xcos(x)

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ddx (2x+xsin(x))

(f±g)′=f ′±g′ :استعمل قانون الجمع

=ddx (2x)+ddx (xsin(x))

ddx (2x)=2

ddx (xsin(x))=sin(x)+xcos(x)

=2+sin(x)+xcos(x)

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

inverse of 20((x-3)(x+4))/((x-3)^2(x-5))

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