Question regarding differentiation of x with respect to x

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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!
 
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Answers and Replies

  • #2
Stephen Tashi
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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
pasmith
Homework Helper
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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?
If you increase [itex]x[/itex] by [itex]\delta x[/itex], then you have increased [itex]x[/itex] by [itex]\delta x[/itex].
 
  • #4
HallsofIvy
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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?
It should be obvious that the "rate of change of x compared to the rate of change of x" is 1!

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!
 
  • #5
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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}##
 
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  • #6
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It should be obvious that the "rate of change of x compared to the rate of change of x" is 1!
This...
 

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