5/(x-2)= 4/(x-1)

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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

simplify solve for inverse tangent line

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ddx (√2x²−1)

Solution

2x√2x²−1

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ddx (√2x²−1)

Apply the chain rule: 12√2x²−1 ddx (2x²−1)

=12√2x²−1 ddx (2x²−1)

ddx (2x²−1)=4x

=12√2x²−1 · 4x

Simplify 12√2x²−1 · 4x: 2x√2x²−1

=2x√2x²−1

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

5/(x-2)= 4/(x-1)

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