5.3 The Fundamental Theorem of Calculus/53 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/53" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T22:24:46Z

Protected "5.3 The Fundamental Theorem of Calculus/53" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 22:24, 6 September 2022
(No difference)

2022-09-06T22:24:30Z

← Older revision		Revision as of 22:24, 6 September 2022
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	\frac{d}{dx}[g(x)] &= \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right] \\[2ex]		\frac{d}{dx}[g(x)] &= \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right] \\[2ex]

	&=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1} ~~\\[2ex]~~		&=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1} =3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1} \\[2ex]

	&=3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1} \\[2ex]

	&=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1} \\[2ex]		&=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1} \\[2ex]

2022-09-06T22:24:06Z

← Older revision		Revision as of 22:24, 6 September 2022
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	\begin{align}		\begin{align}

	\frac{d}{dx}[g(x)] = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]~~=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1}~~ \\[2ex]		\frac{d}{dx}[g(x)] &= \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right] \\[2ex]

	=3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1} \\[2ex]		&=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1} \\[2ex]

	=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1} \\[2ex]		&=3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1} \\[2ex]

			&=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1} \\[2ex]

	\end{align}		\end{align}
	</math>		</math>

2022-09-06T22:23:47Z

← Older revision		Revision as of 22:23, 6 September 2022
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	<math>\~~frac~~{d}~~{dx}[g(x)] = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1}</math>~~		<math>
			\begin{align}

			\frac{d}{dx}[g(x)] = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1} \\[2ex]

	~~<math>~~=3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1}~~</math>~~		=3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1} \\[2ex]

			=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1} \\[2ex]

	~~<math>=~~\~~frac~~{~~3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1~~}</math>		\end{align}
			</math>

2022-09-06T22:22:39Z

← Older revision		Revision as of 22:22, 6 September 2022
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	<math>\frac{d}{dx}g(x) = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1}</math>		<math>\frac{d}{dx}[g(x)] = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1}</math>

2022-09-06T22:22:26Z

← Older revision		Revision as of 22:22, 6 September 2022
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	<math>=3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1}</math>=		<math>=3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1}</math>


	<math>=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1}</math>		<math>=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1}</math>

2022-09-06T22:22:20Z

← Older revision		Revision as of 22:22, 6 September 2022
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	<math>=(3\frac{9x^2-1}{9x^2+1})-(2\frac{4x^2-1}{4x^2+1})</math>=		<math>=3\cdot\frac{9x^2-1}{9x^2+1}-2\cdot\frac{4x^2-1}{4x^2+1}</math>=


	<math>=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1}</math>		<math>=\frac{3(9x^2-1)}{9x^2+1}-\frac{2(4x^2-1)}{4x^2+1}</math>

2022-09-06T22:21:58Z

2022-09-06T22:20:42Z

← Older revision		Revision as of 22:20, 6 September 2022
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	<math>\frac{d}{dx}g(x) = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=(3\cdot\frac{(3x)^2-1}{(3x)^2+1})-(2\cdot\frac{(2x)^2-1}{(2x)^2+1})</math>		<math>\frac{d}{dx}g(x) = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=3\cdot\frac{(3x)^2-1}{(3x)^2+1}-2\cdot\frac{(2x)^2-1}{(2x)^2+1}</math>

2022-09-06T22:20:27Z

← Older revision		Revision as of 22:20, 6 September 2022
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	<math>\frac{d}{dx}g(x) = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=(3\frac{3x^2-1}{3x^2+1})-(2\frac{2x^2-1}{2x^2+1})</math>		<math>\frac{d}{dx}g(x) = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=(3\cdot\frac{(3x)^2-1}{(3x)^2+1})-(2\cdot\frac{(2x)^2-1}{(2x)^2+1})</math>