5.3 The Fundamental Theorem of Calculus/17

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}

${\frac {d}{dx}}(g(x))={\frac {d}{dx}}\left(\int _{1-3x}^{1}{\frac {u^{3}}{(1+u^{2})}}du\right)=(0)\cdot {\frac {(1)^{3}}{(1+(1)^{2})}}-(-3)\cdot {\frac {(-3)^{3}}{(1+(-3)^{2})}}$

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)} }