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> <br><br> | <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)=\frac{-arctan\frac{1}{x}})</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) | |||
=\frac{-\arctan{\frac{1}{x}})</math> <br><br> | |||
<math>\text{Therefore, } g'(x)=</math> | <math>\text{Therefore, } g'(x)=</math> | ||
Revision as of 20:18, 25 August 2022
Failed to parse (syntax error): {\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) =\frac{-\arctan{\frac{1}{x}})}
