5.3 The Fundamental Theorem of Calculus/33: Difference between revisions

Revision as of 19:44, 25 August 2022

$h(x)=\int _{2}^{1/x}\arctan(t)dt$

$h'(x)={\frac {-1}{x^{2}}}/(arctan({\frac {1}{x}})-0$

Revision as of 19:43, 25 August 2022 (view source) Ricardom59381@students.laalliance.org (talk \| contribs) No edit summary ← Older edit		Revision as of 19:44, 25 August 2022 (view source) Ricardom59381@students.laalliance.org (talk \| contribs) No edit summary Newer edit →
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	<math>h(x)=\int_{2}^{1/x}\arctan(t)dt</math>		<math>h(x)=\int_{2}^{1/x}\arctan(t)dt</math>

	<math>h'(x)=\frac{-1}{x^2}(arctan(\frac{1}{x}-0)</math>		<math>h'(x)=\frac{-1}{x^2}/(arctan(\frac{1}{x})-0</math>