5.3 The Fundamental Theorem of Calculus/13: Difference between revisions
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=\frac{-\arctan\frac{1}{x}}{x^2}</math> <br><br> | =\frac{-\arctan\frac{1}{x}}{x^2}</math> <br><br> | ||
<math>\text{Therefore, } | <math>\text{Therefore, } h'(x)=\frac{-\arctan\frac{1}{x}}{x^2}</math> | ||
Latest revision as of 20:15, 6 September 2022
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 \frac{d}{dx}\left[h(x)\right] =\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right] =\frac{-1}{x^2}\cdot\left(\arctan\left(\frac{1}{x}\right)\right)-0\cdot(\arctan\left(2)\right) =\frac{-\arctan\frac{1}{x}}{x^2}}
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, } h'(x)=\frac{-\arctan\frac{1}{x}}{x^2}}