5.3 The Fundamental Theorem of Calculus/15 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/15" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T20:15:45Z

Protected "5.3 The Fundamental Theorem of Calculus/15" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

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Revision as of 20:15, 6 September 2022

2022-09-06T20:12:39Z

← Older revision		Revision as of 20:12, 6 September 2022
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	<math>		<math>
	\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})		\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})
	</math>		</math>

2022-09-06T20:12:26Z

← Older revision		Revision as of 20:12, 6 September 2022
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	<math>		<math>
	\text{Therefore, } y' = \sec^{2}(x)\sqrt{tan(x)+\sqrt tan(x)})		\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})
	</math>		</math>

2022-09-06T20:12:13Z

← Older revision		Revision as of 20:12, 6 September 2022
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	<math>		<math>
	\text{Therefore, } y' = \sec^{2}(x)~~\cdot~~\sqrt{tan(x)+\sqrt tan(x)})		\text{Therefore, } y' = \sec^{2}(x)\sqrt{tan(x)+\sqrt tan(x)})
	</math>		</math>

2022-09-06T20:12:03Z

← Older revision		Revision as of 20:12, 6 September 2022
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	\begin{align}		\begin{align}

	\frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\right]=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})		\frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\right]=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})

	\end{align}		\end{align}

2022-09-06T20:11:42Z

← Older revision		Revision as of 20:11, 6 September 2022
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	\end{align}		\end{align}
	</math>		</math>


	<math>		<math>
	\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})		\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})
	</math>		</math>

2022-09-06T20:11:37Z

← Older revision		Revision as of 20:11, 6 September 2022
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	</math>		</math>

	~~In this problem~~ <math>a^\~~prime~~{(x)~~}= 0</math>, so when it is multiplied by <math>f(a~~(x))</math> ~~it will result in 0 and doesn't need to be added.~~		<math>
			\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})
			</math>

2022-09-06T20:10:59Z

← Older revision		Revision as of 20:10, 6 September 2022
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	\begin{align}		\begin{align}

	\frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\right]=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})		\frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\right]=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})

	\end{align}		\end{align}

2022-09-06T20:10:35Z

← Older revision		Revision as of 20:10, 6 September 2022
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	<math>y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt</math>		<math>y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt</math>


	<math>		<math>

2022-09-06T20:10:25Z

← Older revision		Revision as of 20:10, 6 September 2022
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	\begin{align}		\begin{align}

	\frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\]~~right~~=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})		\frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\right]=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})

	\end{align}		\end{align}