5.3 The Fundamental Theorem of Calculus/15: Difference between revisions
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\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)}) | |||
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Revision as of 20:11, 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 y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt}
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 \begin{align} \frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\right]=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)}) \end{align} }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\text{Therefore, }}y'=\sec ^{2}(x)\cdot {\sqrt {tan(x)+{\sqrt {t}}an(x)}})}