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

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<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)^3}{(1+(1)^2)}
<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)^3}{(1+(1)^2)}


</math>


(0)*f(1)-(-3)*f(1-3x)</math>


which is equal to <math>(3)*f(1-3x)</math>
<math>
 
\text{Therefore, } g'(x) = y^{2}\sin{(y)}
which is=<math>3*\frac{(1-3x)^3}{(1+(1-3x)^2)}</math>
</math>
or simplified to <math>\frac{3*(1-3x)^3}{(1+(1-3x)^2)}</math>
 
 
 
 
 
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Revision as of 20:24, 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 (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}(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)} }


Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\text{Therefore, }}g'(x)=y^{2}\sin {(y)}}

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