5.3 The Fundamental Theorem of Calculus/17: Difference between revisions
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<math>y=\int_{1-3x}^{1}\frac{u^3}{1+u^2} du</math> | |||
<math>y=\ | |||
<math> | |||
\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} | |||
</math> | |||
<math> | |||
\text{Therefore, } y' = \frac{3(1-3x)^3}{1+(1-3x)^2} | |||
</math> | |||
Latest revision as of 20:31, 6 September 2022
