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>


FTC 1:
<math>\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)\,dt=b^\prime{(x)}\cdot\,f(b(x))-\,a^\prime{(x)}\cdot\,f(a(x))</math>


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


\frac{d}{dx}= \int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\\
\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)}})
 
&= \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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<math>
 
\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})
In this problem <math>a^\prime{(x)}= 0</math>, so when it is multiplied by <math>f(a(x))</math> it will result in 0 and doesn't need to be added.
</math>

Latest revision as of 20:15, 6 September 2022



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