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> | |||
<math> | |||
\text{Therefore, } g'(x) = y^{2}\sin{(y)} | |||
</math> | |||
Revision as of 20:24, 6 September 2022