Why do we use"dx" in the derivative dy/dx?

[sviego]

Homework Statement

What purpose does "dx" have, and how can one solve for it?

I understand how to solve derivatives, but am confused as to why we divide by "dx" when we're clearly only solving for the derivative of "dy." Thus, couldn't one write d/dx(x^2) as simply d(x^2)?

Relevant Equations

dx = ?

My theory is that dx = 1(x^0) = 1, which would mean d/dx(x^2) = 2(x^1)/1(x^0) = 2x/1 = 2x.

I know that the derivative is literally the change in "y" over change in "x," but am confused as to what value the change in "x" has.

 

[Orodruin]

It does not have a value. It is a notation for a particular limit, i.e., the entire $dy/dx$ should be seen as short-hand for $\lim_{\epsilon \to 0}[(y(x+\epsilon)-y(x))/\epsilon]$.

 

[jtbell]

the entire $dy/dx$ should be seen as short-hand for $\lim_{\epsilon \to 0}[(y(x+\epsilon)-y(x))/\epsilon]$.

Or, changing your notation a bit: define $\Delta x = (x+ \epsilon) - x = \epsilon$ and $\Delta y = y(x+\epsilon)-y(x)$. Then $\frac {dy} {dx}$ becomes a short-hand for $$\lim_{\Delta x \to 0} \frac {\Delta y} {\Delta x}$$ The notations $dx$ and $dy$ are supposed to make you think about infinitesimally tiny $\Delta x$ and $\Delta y$.

 

[Stephen Tashi]

Homework Statement: What purpose does "dx" have, and how can one solve for it?

To repeat what others said, the modern interpretation of $dy/dx$ in a calculus course does not regard $dy$ and $dx$ as being numbers and it does not regard $dy/dx$ as being a fraction involving two numbers. If f(x) is a function, perhaps you have seen the notation f'(x) to denote the derivative of f. The modern interpretation of $df/dx$ is that it means $f'(x)$.

There are advanced mathematical subjects where notation such $df$ is used without any denominator and there is an advanced (but not particularly popular) method of inventing a number system which contains "infinitesimals" and these are often denoted by symbols like $dx$.

Calculus has a long history and the early inventors of calculus did not have the modern concept of limits. So they did think of $dy/dx$ as fraction.

My theory is that dx = 1(x^0) = 1, which would mean d/dx(x^2) = 2(x^1)/1(x^0) = 2x/1 = 2x.

I don't know what you mean by "d/dx(^2)". For example, is "d/dx(^2) sin(x)" supposed to have a meaning?

Before the invention of limits, the approach to derivatives went something like this:
Let $y = x^2$. The derivative of y (in the old days) was $\frac{dy}{dx} = \frac{(x+dx)^2 - x^2}{dx} = \frac{x^2 + 2xdx + (dx)^2 - x^2}{dx} = \frac{ 2xdx + (dx)^2}{dx} = 2x + dx$. To get the final result $\frac{dy}{dx} = 2x$ we are forced to think of $dx$ as a number that is effectively zero, but not causing the embarrasement of dividing by zero before we reduce the fraction. There is also the difficulty that the numerator $(x + dx)^2 - x^2$ would be zero if $dx$ was zero.

There was a struggle to interpret the above algebraic manipulations in various ways. A satisfactory method of interpreting symbols like $dx$ as a new type of number was finally developed in the 1960's . ( https://en.wikipedia.org/wiki/Non-standard_analysis ). However, long before that, the concept of limits was introduced to define the concepts of calculus in a precise manner. Typical modern calculus texts define concepts in terms of limits, not in terms of nonstandard analysis.

Notation such as $\frac{dy}{dx}$ is useful in remembering certain rules of calculus and many physics texts present arguments treating $dy$ and $dx$ as numbers. Thinking about them this way is useful in understanding physics.

For example, the "chain rule" is $D_x f(g(x)) = f'(g(x)) g'(x)$. This can be remembered by $\frac{df}{dx} = \frac{dy}{dg} \frac{dg}{dx}$, which treats the derivatives as fractions. However, it might lead to making the mistake $D_x sin(x^2) = cos(x) 2x$. The other notation makes it clear that $f'()$ is to be evaluated at $x^2$, so we get $D_x sin(x^2) = cos(x^2) 2x$.

 

[sviego]

Wow, this was really informative! Thank you Stephen.

I don't know what you mean by "d/dx(^2)".

I meant "the derivative of x(^2)," but as a random example. I see now, however, that we're not trying to find the derivative of x because (and thanks to limits) the denominator for dx is simply zero.

The history of derivatives without limits is fascinating! I forget that calculus wasn't born in a single day; it's constantly shaping and evolving.

 

[sviego]

Hi Orodruin! This is my first time writing on these forums, so thanks for the help.

It does not have a value. It is a notation for a particular limit, i.e., the entire $dy/dx$ should be seen as short-hand for $\lim_{\epsilon \to 0}[(y(x+\epsilon)-y(x))/\epsilon]$.

I see, so "dx" is simply shorthand for the denominator of the limit where dx→0 when dx = h and h→0.

 

[sviego]

Hi jtbell! What an interesting addition to Orodiun's response, I really appreciate it.

The notations $dx$ and $dy$ are supposed to make you think about infinitesimally tiny $\Delta x$ and $\Delta y$.

That makes so much sense! Simply put, the derivative describes the microscopic change in y and x as x approaches zero.

 

[PeroK]

Hi Orodruin! This is my first time writing on these forums, so thanks for the help.
I see, so "dx" is simply shorthand for the denominator of the limit where dx→0 when dx = h and h→0.

$h$ is a number. In $\frac{dy}{dx}$ then $dx$ has no independent meaning. The whole thing means "the derivative of $y$ with respect to $x$'. But $dy$ and $dx$ have no independent meaning.

Things in mathemetics are what you define them to be.

One of the next steps in calculus is to give meaning to $dx$ as a "differential" or "infinitesimal". But, technically, that is different from the use of $dx$ in the derivative.

 

[PeroK]

Hi jtbell! What an interesting addition to Orodiun's response, I really appreciate it.
That makes so much sense! Simply put, the derivative describes the microscopic change in y and x as x approaches zero.

As this is "calculus and beyond", "microscopic" has no mathematical meaning. The derivative is a limit, which has a well-defined mathematical meaning. The definition of a limit uses only the properties of real numbers.

 

[Mark44]

My theory is that dx = 1(x^0) = 1, which would mean d/dx(x^2) = 2(x^1)/1(x^0) = 2x/1 = 2x.

I don't know what you mean by "d/dx(^2)". For example, is "d/dx(^2) sin(x)" supposed to have a meaning?

Stephen, the OP didn't write "d/dx(^2)." You quoted what the OP wrote, but apparently misread it.

 

[WWGD]

Ultimately, the expression f'(x)dx is called the differential* and measures the (linear) approximation to the change of the values of the function. If, say, f(x)=$x^2$ then the differential. 2xdx measures changes in a neighborhood (x, x+h) or ( x, x+ dx). Maybe we can ask Freshmeister @fresh_42 to chime in?

* Different authors have different notation.