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I have a question below but I missed this day of class maybe someone can show me how to approach?

Find $dy$ and evaluate $dy$ for the given values of $x$ and $dx$

  1. $\displaystyle y=e^\frac{x}{10}$
  2. $\displaystyle x = 0$ and $dx = 0.1$
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    Well, it seems like you should recall $ \mathrm{d}y=\frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x$ From there, it should be straightforward to solve...2012-11-23

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With differentials, it's an abuse of notation, but the answer is that $ \mathrm{d}y=\frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x$ Since $\mathrm{d}x$ is impossible to use, we make the approximation $\Delta x\approx\mathrm{d}x$.

For us, $\Delta x=0.1$. Now, we find the derivative $\frac{\mathrm{d}}{\mathrm{d}x}e^{x/10}=\frac{1}{10}e^{x/10}.$ Then we plug in the data: $\mathrm{d}y\approx \left.\left(\frac{1}{10}e^{x/10}\right)\Delta x\right|_{x=0}$ which means we are evaluating the parenthetic term when $x=0$, and multiply by $\Delta x=0.1$.

Now we have to just plug these things in to find: $\begin{align}\left.\left(\frac{1}{10}e^{x/10}\right)\Delta x\right|_{x=0}&=\left(\frac{1}{10}e^{0}\right)0.1\\ &=0.01\end{align}$ Thus $\mathrm{d}y\approx 0.01$.

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    @AlexNelson, you're welcome. Also, thanks to robjohn for understanding arithmetic. It constantly eludes me, and I find myself running to fields, rings, and groups for comfort. :P2012-11-23