5.3 The Fundamental Theorem of Calculus/17: Difference between revisions

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<math>
<math>
\frac{d}{dx}(g(x))=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) = (0)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}
\frac{d}{dx}(g(x))=\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{(-3)^3}{(1+(-3)^2)}
-(-3)\cdot\frac{(1-3x)^3}{(1+(1-3x)^2)}


</math>
</math>

Revision as of 20:26, 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 g(x)=\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}}(g(x))={\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, } g'(x) = y^{2}\sin{(y)} }