5.3 The Fundamental Theorem of Calculus/11

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Revision as of 19:24, 25 August 2022 by Elizabethh58413 (talk | contribs) (Created page with "<math> g(x)= \int_{\pi}^{x}\sqrt{1+sec(t)}\cdot dt </math> <br><br> <math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <br><br> <math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>")
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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 g(x)= \int_{\pi}^{x}\sqrt{1+sec(t)}\cdot 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 \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}}

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, } g'(x)=-\sqrt{1+sec(x)}}

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