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amplitude of arccos(x)

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

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

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ddx (x¹⁴4−x¹⁷ )

Solution

56x¹³+3x³⁰(4−x¹⁷)²

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ddx (x¹⁴4−x¹⁷ )

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

=ddx (x¹⁴)(4−x¹⁷)−ddx (4−x¹⁷)x¹⁴(4−x¹⁷)²

ddx (x¹⁴)=14x¹³

ddx (4−x¹⁷)=−17x¹⁶

=14x¹³(4−x¹⁷)−(−17x¹⁶)x¹⁴(4−x¹⁷)²

Simplify 14x¹³(4−x¹⁷)−(−17x¹⁶)x¹⁴(4−x¹⁷)² : 56x¹³+3x³⁰(4−x¹⁷)²

=56x¹³+3x³⁰(4−x¹⁷)²

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

amplitude of arccos(x)

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