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

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<math>g(x)=\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du</math>
<math>y=\int_{1-3x}^{1}\frac{u^3}{1+u^2} du</math>




<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)^3}{(1+(1)^2)}
\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)}
-(-3)\cdot\frac{(1-3x)^3}{1+(1-3x)^2}


</math>
</math>
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<math>
<math>
\text{Therefore, } g'(x) = y^{2}\sin{(y)}
\text{Therefore, } y' = \frac{3(1-3x)^3}{1+(1-3x)^2}
</math>
</math>

Latest revision as of 20:31, 6 September 2022



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