Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
See also this previous question; it may help. –  Arturo Magidin Oct 4 '11 at 16:09

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
So h is the difference between x and a? Also in the example I have no idea how to do it, how to set it up or how to get the derivative of e to the x/10 –  user138246 Oct 4 '11 at 16:35
    
Gortaur explains pretty well how to construct your answer using the definition of the derivative. You can also use the rules of differentiation (which are just shortcuts to finding the above). In this case, you would need to know that the derivative of $e^x$ is $e^x$ and that the derivative of $f(g(x))$ is $f'(g(x))g'(x)$ (the chain rule). –  kbolino Oct 4 '11 at 18:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.