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>\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}</math> <br><br> | |||
<math>\text{Therefore, } h'(x)=\frac{-\arctan\frac{1}{x}}{x^2}</math> | |||
Latest revision as of 20:15, 6 September 2022
![{\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}}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/38499f58d730e54a4a6dbeeda4924be33b09cae5)
