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I don't know what is going on at all.

I am supposed to find the differential of each function. It was explained thoroughly to me for about 2 hours but I can't figure it out. Stewart is amazing at doing that. The first problem he wants me to solve sure is easy: $y=e^{x/10}$, $x=0$, $dx=.1$. He gave me a formula of $dy=f'(x) \, dx$ I don't really know what that is but it doesn't seem to work when I try to use it.

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1 Answer

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It depends on which course are you following, but the most basic explanation. Given a differentiable function $f$ on the real line, a point $x$ and an increment $h$ you would like to find how will the value of $f$ change from $f(x)$ to $f(x+h)$. Clearly, the good way to compare them is to subtract: $$ \Delta f(x,h) = f(x+h) - f(x). $$ For example, if $f = kx$ then $\Delta f(x,h) = k(x+h)-kx = kh$.

If $f$ has more complicated form, the structure of its increment $\Delta f(x,h)$ could be weird. On the other hand, if the increment $h$ is small - you can still expect kind linear behavior of $\Delta f$. You decompose $\Delta f$ into two terms - the main (linear) one and the rest which should be very 'small': $$ \Delta f(x,h) = k(x)\cdot h+\varepsilon(h)h\quad(\star) $$ where $\varepsilon(h)\to 0$ with $h\to 0$, so it is negligible. How to find such a value $k(x)$ at $x$? Ok, we know that $\varepsilon(h)\to 0$ so we should just extract $\varepsilon(h)$ from $(\star)$ and calculate the limit: $$ \varepsilon(h) = \frac{\Delta f(x,h)}{h}-k\to0, $$ so $$ k = \lim\limits_{h\to 0}\frac{\Delta f(x,h)}{h} = \lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h} = f'(x). $$ The part $k(x)\cdot h = f'(x)h=:\mathrm df(x,h)$ is called the differential of $f$ at the point $x$ and increment $h$. Note that $\mathrm df(x,h)$ is a function both of a point and the increment.

Now about your example: $f'(x) = \frac1{10}e^{x/10}|_{x=0} = \frac1{10}$. The increment denoted as $\mathrm dx (= h) = 0.1$, so $$ \mathrm df(x,dx) = \frac1{10}\cdot 0.1 = \frac1{100}. $$

For more information you may wanna read the Wikipedia page.

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