derivative of (4x/(5x^2+3))

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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

simplify solve for inverse tangent line

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laplace transform cosh²(x)

Solution

12s +s2(s²−4)

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Solution steps

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L{cosh²(x)}

Use the following identity: cosh²(x)=12 +cosh(2x)12

=L{12 +cosh(2x)12 }

Use the linearity property of Laplace Transform:
For functions f(t), g(t) and constants a, b: L{a·f(t)+b·g(t)}=a·L{f(t)}+b·L{g(t)}

=L{12 }+12 L{cosh(2x)}

L{12 }: 12s

L{cosh(2x)}: ss²−4

=12s +12 · ss²−4

Refine 12s +12 · ss²−4 : 12s +s2(s²−4)

=12s +s2(s²−4)

Solve instead

inverse laplace cosh²(x)

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

derivative of (4x/(5x^2+3))

Solution

Solution steps

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