5.3 The Fundamental Theorem of Calculus/7: Difference between revisions
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<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]= | <math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]= | ||
(1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math><Br> | (1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math><Br> | ||
<br> | |||
<math>\text{Therefore, } g'(x)=\frac{1}{x^3+1}</math> | <math>\text{Therefore, } g'(x)=\frac{1}{x^3+1}</math> | ||
Revision as of 19:49, 6 September 2022
![{\displaystyle {\frac {d}{dx}}\left[g(x)\right]={\frac {d}{dx}}\left[\int _{1}^{x}{\frac {1}{t^{3}+1}}dt\right]=(1)\left({\frac {1}{(x)^{3}+1}}\right)-(0)\left({\frac {1}{(1)^{3}+1}}\right)={\frac {1}{x^{3}+1}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6cf243c7869bd94ab3b91c68f55b995a0942c27d)
