5.3 The Fundamental Theorem of Calculus/13: Difference between revisions
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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> <br><br> | ||
<math>\frac{d}{dx}\left[h(x)\right]=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]=\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)=\arctan(\frac{1}{x})</math> | <math>\frac{d}{dx}\left[h(x)\right]=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]=\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)=\arctan(\frac{1}{x})</math> <br><br> | ||
<math>\text{Therefore, } g'(x)=</math> | <math>\text{Therefore, } g'(x)=</math> | ||
Revision as of 20:14, 25 August 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}\left[h(x)\right]={\frac {d}{dx}}\left[\int _{2}^{1/x}\arctan(t)dt\right]={\frac {-1}{x^{2}}}\cdot (\arctan \left({\frac {1}{x}}\right))-0\cdot (\arctan \left(2)\right)=\arctan({\frac {1}{x}})}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \text{Therefore, } g'(x)=}