# Question regarding differentiation of x with respect to x

1. Feb 8, 2015

### JDC123

Hi everybody,

I'm having a little difficulty understanding the differentiation of x with respect to x. When a function, f(x) is differentiated, each term is differentiated with respect to x, correct? So, when differentiating y=x, we would have d(y)/dx = d(x)/dx. To my (very limited) knowledge, dy/dx can be stated as the observed change in y for a given change in x (please correct me if I'm wrong). Applying this same logic, dx/dx can be stated as an observed change in x for a given change in x. This is where I'm hung up; how is it possible to compare the rate of change of x to itself?

My intuition (which is obviously faulty) tells me that in order to determine a rate of change, a second variable (in this case, y) must be involved.

I'd greatly appreciate any insight as I'm missing something quite trivial.
Thanks!

Last edited: Feb 8, 2015
2. Feb 8, 2015

### Stephen Tashi

You are using intuitive and imprecise notions. The derivative of a function f(x) is defined as a limit. Look at what the actual definition of derivative tells you about differentiating the function f(x) = x.

To some extent it is possible to reason intuitively with symbols like dx and dy, but it takes experience to know what sort of intuitive reasoning actually works.

3. Feb 9, 2015

### pasmith

If you increase $x$ by $\delta x$, then you have increased $x$ by $\delta x$.

4. Feb 11, 2015

### HallsofIvy

It should be obvious that the "rate of change of x compared to the rate of change of x" is 1!

5. Mar 1, 2015

### foobar99

Think of it as rate of change of one thing w.r.t another.

So, if I get paid P=$2 per foot of ditch, and I dig S=3 feet per day, $dP/dt = 2, dS/dP = 3$, and $dP/dt = 6$ (i.e.,$6 per day). The last one, by the way, is the chain rule.

So, loosely speaking, $dP/dP = 1$ - i.e., I get one dollar for each dollar I get. And $dS/dS$ also equals 1 - I dig one foot for every foot I dig. Trivial and sloppy but true.

The notation is endlessly confusing, I would agree. Wait til you find yourself doing things like figuring the derviative of something with respect to another thing. For example, if$g = g(t), and f = f(q)$, and we then look at f(g), which we call f(g(t)), it gets mighty confusing to take the derivative of f w.r.t. g, $df/dg.$ if $g = t^2$, you can find yourself taking $df/dt^2$. In fact, if $g(t) = e^{t^2}$, it is perfectly reasonable to find $dg/de^{t^2}$

Last edited: Mar 1, 2015
6. Mar 1, 2015

This...