5.3 The Fundamental Theorem of Calculus/15: Difference between revisions

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\frac{d}{dx}
\frac{d}{dx}


&= \int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt  
&= \int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt  

Revision as of 19:53, 25 August 2022

Use part 1 of the FTC to find the derivative of the function: 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}

FTC 1: 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}\int_{a(x)}^{b(x)}f(t)\,dt=b^\prime{(x)}\cdot\,f(b(x))-\,a^\prime{(x)}\cdot\,f(a(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 \begin{align} \frac{d}{dx} &= \int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt &= \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} }



In this problem 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 a^\prime{(x)}= 0} , so when it is multiplied by 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 f(a(x))} it will result in 0 and doesn't need to be added.