5.3 The Fundamental Theorem of Calculus/15: Difference between revisions
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\begin{align} | \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)}) | \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} | ||
</math> | </math> | ||
<math> | |||
\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}}) | |||
</math> | |||
Latest revision as of 20:15, 6 September 2022
![{\displaystyle {\begin{aligned}{\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{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/dc58a595f2aad3e2c34089e3ccadae24a38c442f)
