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domain of 2sin(5x)

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

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ddx

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

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ddx (23−4x )

Solution

8(3−4x)²

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Solution steps

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ddx (23−4x )

Take the constant out: (a·f)′=a·f ′

=2ddx (13−4x )

Apply exponent rule: 1a =a⁻¹

=2ddx ((3−4x)⁻¹)

Apply the chain rule: −1(3−4x)² ddx (3−4x)

=−1(3−4x)² ddx (3−4x)

ddx (3−4x)=−4

=2(−1(3−4x)² (−4))

Simplify 2(−1(3−4x)² (−4)): 8(3−4x)²

=8(3−4x)²

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

domain of 2sin(5x)

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