5.3 The Fundamental Theorem of Calculus/17: Difference between revisions
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\frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{1+u^2} du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)} | \frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{1+u^2}\,du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)} | ||
-(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)} | -(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)} | ||
Revision as of 20:30, 6 September 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=\int _{1-3x}^{1}{\frac {u^{3}}{1+u^{2}}}du}
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}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{1+u^2}\,du\right) = (0)\cdot\frac{(1)^3}{(1+(1)^2)} -(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)} }
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, } y' = \frac{3(1-3x)^3}{1+(1-3x)^2} }