5.3 The Fundamental Theorem of Calculus/15: Difference between revisions

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\begin{align}
\begin{align}


\frac{d}{dx}(y)= \int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})
\frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\]right=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})


\end{align}
\end{align}

Revision as of 20:10, 6 September 2022

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\]right=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)}) \end{align} }

In this problem , so when it is multiplied by it will result in 0 and doesn't need to be added.