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What does $dy = f'(x)dx$ I don't get the meaning of this or how can I apply because is not $f'(x)$ the instantaneous change or maybe I am not understanding what $dx$ and $dy$ really means

kprincipe
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  • $f(x + \varepsilon) = f(x) + \varepsilon f'(x)$ –  Apr 12 '16 at 02:26
  • Related: http://math.stackexchange.com/questions/23902/what-is-the-practical-difference-between-a-differential-and-a-derivative, http://math.stackexchange.com/questions/200393/what-is-dx-in-integration – Hans Lundmark Apr 12 '16 at 07:23

3 Answers3

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The easiest way to understand differentials intuitively is as infinitesimal distances, with derivatives being true fractions, or ratios, of those distances.

The expression $dy = f'(x)dx$ is equivalent to $dy = \frac{dy}{dx}dx$, and $f'(x)$ or $\frac{dy}{dx}$ is literally the slope of the line tangent to the function at any point. Thus, the derivative multiplied by an infinitesimal length dx is equal to an infinitesimal change in the y value. You can imagine that for every unit length you move in the x direction, y increases by $f'(x)$ times that length. Now most functions aren't linear (and so most derivatives aren't constant), and so adding up those values is more difficult. That's why the integral is useful. You can add up all of the small $dy$ values over the entire function to get the function itself. That's the fundamental theorem of calculus.

$\int_{a}^{b}\frac{dy}{dx}dx = f(x)$

It really helps to think of differentials as small distances that can be compared. Then an expression like

$dy = 3dx$

makes a lot more sense. $dy$ is literally 3 times longer than $dx$, meaning that every small change in the x value increases the y value by 3 times that much. This is an important concept, so feel free to ask questions.

JAustin
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Note that $\Delta f(x)= f'(x)\Delta x + \epsilon (\Delta x) \Delta x$ as well as that $dx= \Delta x$. Where $\epsilon(\Delta x) \rightarrow 0$ as $\Delta x \rightarrow 0$ . My understanding is that $dy$ is something of an approximation of the change in $f(x)$ for since small enough $\Delta x $ implies $dy \approx \Delta f(x)$. It is possible then to think of $dy$ as somewhat of an approximation to the change in f of x. Lastly manipulating the equation you said results in the notation of $ \frac{dy}{dx} = f'(x)$.Hope this helps.

A.Riesen
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You must understand that $dy = f'(x)dx$ implies that $y = f(x)$. So if you plot the curve of $y$ vs. $x$, then the derivative gives you the slope of the line which is tangent to the curve at each point being evaluated. (This of course assuming that $y$ is continuous.)

So $dy$ is the change in $y$ for the corresponding change in $x$ which is $dx$.

MaxW
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