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}{ | \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}{ | -(-3)\cdot\frac{(1-3x)^3}{1+(1-3x)^2} | ||
</math> | </math> | ||
Revision as of 20:30, 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_{1-3x}^{1}\frac{u^3}{1+u^2} du}
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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} }