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inverse of f(x)=x(sqrt((1+2/x))-1)

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

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ddz ((z−8)²z²−64 )

Solution

16(z+8)²

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ddz ((z−8)²z²−64 )

Apply the Quotient Rule: (fg )^′=f ′·g−g′·fg²

=ddz ((z−8)²)(z²−64)−ddz (z²−64)(z−8)²(z²−64)²

ddz ((z−8)²)=2(z−8)

ddz (z²−64)=2z

=2(z−8)(z²−64)−2z(z−8)²(z²−64)²

Simplify 2(z−8)(z²−64)−2z(z−8)²(z²−64)² : 16(z+8)²

=16(z+8)²

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

inverse of f(x)=x(sqrt((1+2/x))-1)

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