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

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Use part 1 of the FTC to find the derivative of the function:
 
<math>y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt</math>
<math>y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt</math>


<math>
<math>
\begin{align}
\begin{align}


y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt =\sqrt{tan(x)+\sqrt tan(x)}\cdot\sec^{2}(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}
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FTC 1:
<math>
<math>\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)\,dt=b^\prime\cdot\,f(b(x))-\,a^\prime\cdot\,f(a(x))</math>
\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})
</math>

Latest revision as of 20:15, 6 September 2022



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