5.3 The Fundamental Theorem of Calculus/17 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/17" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T20:31:58Z

Protected "5.3 The Fundamental Theorem of Calculus/17" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

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Revision as of 20:31, 6 September 2022

2022-09-06T20:30:42Z

← Older revision		Revision as of 20:30, 6 September 2022
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	<math>		<math>
	\frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{1+u^2}\,du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}		\frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{1+u^2}\,du\right) = (0)\cdot\frac{(1)^3}{1+(1)^2}

	-(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}		-(-3)\cdot\frac{(1-3x)^3}{1+(1-3x)^2}

	</math>		</math>

2022-09-06T20:30:25Z

← Older revision		Revision as of 20:30, 6 September 2022
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	<math>		<math>
	\frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{1+u^2} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}		\frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{1+u^2}\,du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}

	-(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}		-(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}

2022-09-06T20:30:15Z

← Older revision		Revision as of 20:30, 6 September 2022
Line 3:		Line 3:

	<math>		<math>
	\frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}		\frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{1+u^2} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}

	-(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}		-(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}

2022-09-06T20:30:02Z

← Older revision		Revision as of 20:30, 6 September 2022
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	<math>~~g(x)~~=\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du</math>		<math>y=\int_{1-3x}^{1}\frac{u^3}{1+u^2} du</math>


	<math>		<math>
	\frac{d}{dx}(~~g(x)~~)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}		\frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}

	-(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}		-(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}
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	<math>		<math>
	\text{Therefore, } g'~~(x)~~ = \frac{3(1-3x)^3}{1+(1-3x)^2}		\text{Therefore, } y' = \frac{3(1-3x)^3}{1+(1-3x)^2}
	</math>		</math>

2022-09-06T20:27:47Z

← Older revision		Revision as of 20:27, 6 September 2022
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	<math>		<math>
	\text{Therefore, } g'(x) = ~~(3)\cdot~~\frac{(1-3x)^3}{1+(1-3x)^2}		\text{Therefore, } g'(x) = \frac{3(1-3x)^3}{1+(1-3x)^2}
	</math>		</math>

2022-09-06T20:27:29Z

← Older revision		Revision as of 20:27, 6 September 2022
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	<math>		<math>
	\text{Therefore, } g'(x) = y^{2}~~\sin~~{(y)}		\text{Therefore, } g'(x) = (3)\cdot\frac{(1-3x)^3}{1+(1-3x)^2}
	</math>		</math>

2022-09-06T20:26:59Z

← Older revision		Revision as of 20:26, 6 September 2022
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	<math>		<math>
	\frac{d}{dx}(g(x))=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1~~-3x~~)^3}{(1+(1~~-3x~~)^2)}		\frac{d}{dx}(g(x))=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}

	-(-3)\cdot\frac{(-3)^3}{(1+(-3)^2)}		-(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}

	</math>		</math>

2022-09-06T20:26:27Z

← Older revision		Revision as of 20:26, 6 September 2022
Line 3:		Line 3:

	<math>		<math>
	\frac{d}{dx}(g(x))=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}		\frac{d}{dx}(g(x))=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}

	-(-3)\cdot\frac{(-3)^3}{(1+(-3)^2)}		-(-3)\cdot\frac{(-3)^3}{(1+(-3)^2)}

2022-09-06T20:25:50Z

← Older revision		Revision as of 20:25, 6 September 2022
Line 4:		Line 4:
	<math>		<math>
	\frac{d}{dx}(g(x))=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}		\frac{d}{dx}(g(x))=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)}

			-(-3)\cdot\frac{(-3)^3}{(1+(-3)^2)}

	</math>		</math>