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I have a fundamental query about the way derivatives can be used in algebraic manipulations.

Say $\dfrac{d(\ln x)}{dx}=\dfrac{1}{x}$

Apparently, this can be manipulated to $d(\ln x)=\dfrac{dx}{x}$. I understand that this can be integrated back to the first equation.

But, the reason this was done was to depict $\dfrac{dx}{x}$ as a percentage change in $x$.

The definition of $\dfrac{df(x)}{dx}=\lim_\limits{h\to 0}\dfrac{f(x+h)-f(x)}{h}$. So, isn't $\dfrac{d}{dx}$ more like a function than a ratio of two quantities ?

If so, how is moving $dx$ to RHS possible as done above ?

Please advise.

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  • $\begingroup$ You are right. $\frac{d}{dx}$ acts as a function. It is mistaken to break up the differentials like that when it's really a notational tool more than anything. What's really going on is that you are evaluating $$\int\frac{d}{dx}\ln x\,dx$$ which is the same as integrating $\frac{1}{x}$ since the two functions are the same. It happens to work out but usually it is a matter of chain rule and the fundamental theorem of calculus that is really going on behind the scenes. $\endgroup$ Mar 5, 2016 at 18:50

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Mathematically speaking, $\frac{d}{dx}$ is an operator whose work is to differentiate functions, just like $+,-,*,/$ are all operators whose respective functions are to add, subtract, multiply and divide.

The multiplication of $dx$ however proceeds as follows: $$\frac{d}{dx}f(x)=f'(x) \Rightarrow \frac{d}{dx}f(x) dx=f'(x) dx \Rightarrow d\{f(x)\}=f'(x) dx$$

Now it is logically deduced with the help of infinitesimals that $$\frac{d}{dx}f(x) dx=d\{f(x)\}$$ It is not that the $dx$ in the numerator and denominator just cancel out like a ratio.

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  • $\begingroup$ $\ldots\,$provided one follows the conventions and definitions that have become the most usual ones since some time in the 19th century. $\qquad$ $\endgroup$ Mar 5, 2016 at 18:57
  • $\begingroup$ Thanks .. but I did not understand the part "it is logically deduced with the help of infinitesimals that $$\frac{d}{dx}f(x) dx=d\{f(x)\}$$" .. any place I can read more on this.. $\endgroup$
    – square_one
    Mar 9, 2016 at 18:00
  • $\begingroup$ @square_one The L.H.S. means "small change in $f(x)$ for unit change $dx$ and then the quantity is multipied by $dx$ to get the net change in $f(x)$" which is the total change in $f(x)$ or as written in R.H.S. $d\{f(x)\}$. $\endgroup$ Mar 10, 2016 at 5:06
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Intuitively, $dx$ is an infinitely small change in the value of $x$. This results in an infinitely small change in $\log x$, and that change is $d(\log x) = \dfrac{dx} x$.

If we then go on to conclude that $$ \int_{[a,b]} g \cdot \log' =\int_a^b g(x) \log' x\,dx = \int_a^b \frac{g(x)} x\,dx $$ then the identity is correct, and essentially that's an application of the chain rule.

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I find it most beneficial to view $\frac{d}{dx}$ as being two operations - a differential and a division. Most of Calculus notation becomes much more clear when you think of it that way. Then the terms $dx$ and $dy$ can be manipulated as algebraic terms.

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I have gone through the same problem when I was learning calculus for the first time. Let me explain how I understood the differential sign $\frac{d}{dx}$. basically the basic concept I understood was;

derivatives are nothing but SLOPES of a certain function.

I started with the definition of slopes which is $\frac{\Delta y}{\Delta x }$, assuming my $\Delta$ simply to mean "significant change or big change in some value". So, the slope of $f(x)$ at some $x$ is simply, $$\frac{\Delta f(x)}{\Delta x} .$$ I asked myself: "can I consider this as a ratio and simply manipulate them as a ratio ($\frac{a}{b}=1\Rightarrow a=b$). I came to the conclusion that I could do so by taking the example of $f(x)$ being a straight line (we use this method if two points are given and we need to find the equation of $y$).

Now that I did this I came back to $\frac{d}{dx}$. So the definition states that $$\frac{df(x)}{dx}=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.$$ So this came from the fact that $h=x_{1}-x$, where $x_{1}$ is a variable and $x$ is some constant we want to find the slope at. Replacing we get, $$\lim_{x_{1}\rightarrow x}\frac{f(x_{1})-f(x)}{x_1-x}.$$ This is the first slope I discussed, it's just that $x_{1}$ and $x$ are so close to each other, so $\frac{df(x)}{dx}$ simply meant slope with $d$ denoting small change or insignificant change. Since I understood that it is nothing but slope which by definition is ratio I understood I could manipulate them.

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