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January 30th, 2016, 10:33 AM   #1
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Can a variable be both Independent and Dependent (Troubles with the Chain Rule)

According to Keisler's Elementary Calculus: an infinitesimal approach
Chapter 2 / Section 2.6 / Pages 88-92
( https://www.math.wisc.edu/~keisler/calc.html )

We have two cases for the Chain Rule, which are: (page 89)

Case 1: dy / dt = dy / dx * dx / dt

where X is independent in dy/dx and t is independent both in dy/dt and dx/dt.

It SEEMS to me that X here, is independent to y, and dependent on t.

Can a variable be dependent and independent at the same time ?

Case 2: dy / dx = (dy/dt) / (dx/dt)

where t is the only independent variable.

If that is the case, then what is the relation between x and y called ? In other words, how can y be a function of x here when both x and y are dependent ?

And finally, if y = f(x) has an inverse function x = g(y), is x independent in both the original function and its inverse ?
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January 30th, 2016, 10:49 AM   #2
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Yes, a variable can be an "independent variable" in one function, and a "dependent variable" in another. In the case where $\displaystyle y= f(x)$ and $\displaystyle x= f^{-1}(y)$, x is the "independent variable" in the first equation and the dependent variable in the second. y is the dependent variable in the first equation and the independent variable in the second. That is one reason why the terms "independent variable" and "dependent variable" are not very useful!
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January 30th, 2016, 11:57 AM   #3
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Case 1:
$y$ is expressed as a function of $x$. Also $x$ can be expressed as a function of $t$. Then, by replacing each occurrence of $x$ in the function $y$ by $x(t)$, the expression in $t$, we have expressed $y$ as a function of $t$. We can determine the rate of change of $y$ with respect to $t$. And it turns out that ${\mathrm d y \over \mathrm d t}={\mathrm d y \over \mathrm d x}{\mathrm d x \over \mathrm d t}$. Very often we then replace each occurrence of $x$ in ${\mathrm d y \over \mathrm d x}$ with $x(t)$.

Case 3:
Dependent/independent? We can graph $y=f(x)$ to get a curve. If we represent our location on the curve with a point $P\big(x,f(x)\big)$, we can imagine moving our point around. If we drag it to the right, we force $x$ to increase. But since the point is constrained to the curve, it will also move up or down. The $y$ coordinate of $P$ depends on the $x$ coordinate.

Conversely, we might pull the point upwards, forcing $y$ to change. Still the point is constrained to the curve, so it also moves to the left or the right. Now the $x$ coordinate of $P$ depends on how the $y$ coordinate changes.

This inter-relation movement can be expressed by the idea of the inverse function. $x=f^{-1}(y)$. Neither $x$ nor $y$ is strictly dependent on the other, but if we are influencing one variable, it is useful to see how that affects the other. We do this by expressing it in terms of the variable we are changing directly.

It turns out that ${\mathrm d x \over \mathrm d y}={1 \over {\mathrm d y \over \mathrm d x}}$.

Case 2:
Sometimes we are able to express both $x$ and $y$ in terms of a third variable $t$. Plotting points $\big(x(t),y(t)\big)$, we still get a curve. So as before we can pull a point on the line to the left or the right and see how it moves up and down as a result. This relationship can be seen more clearly if we rearrange one of the equations $x=g(t)$ to get $t=g^{-1}(x)$. Now we are back to case 1, where we have ${\mathrm d y \over \mathrm d x}={\mathrm d y \over \mathrm d t}{\mathrm d t \over \mathrm d x}$. But case 2 tells us that ${\mathrm d t \over \mathrm d x}={1 \over {\mathrm d x \over \mathrm d t}}$. And thus we have ${\mathrm d y \over \mathrm d x}={{\mathrm d y \over \mathrm d t} \over {\mathrm d x \over \mathrm d t}}$.
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Last edited by skipjack; January 30th, 2016 at 12:37 PM.
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January 30th, 2016, 07:05 PM   #4
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Quote:
Originally Posted by Country Boy View Post
That is one reason why the terms "independent variable" and "dependent variable" are not very useful!
I think the terms "independent variable" and "dependent variable" are useful when trying to get a statistical conclusion, but not when doing something like (dy/dk)(dk/dx) = dy/dx. In some cases it's obvious which variable is independent and which variable is dependent. For example, smoking can cause lung cancer. Getting lung cancer hopefully doesn't cause a nonsmoker to start smoking.
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January 31st, 2016, 05:34 AM   #5
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This problem wasn't about statistics!
Yes, "cause and effect" can be important in applications but there is no "cause and effect" in mathematics.
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January 31st, 2016, 05:45 AM   #6
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I would observe that, although many of us pretend otherwise, statistics is a branch of mathematics.
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January 31st, 2016, 08:16 AM   #7
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A well-known university has a Department of Pure Mathematics and Mathematical Statistics.
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January 31st, 2016, 08:50 AM   #8
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Personally, I don't think about the Leibniz notation when I do chain rule. I imagine something like this in my head

Derivative of f(g(h(x))) is f'(g(h(x))) * g'(h(x)) * h'(x)

This isn't 'correct', but it's enough for me to compute the answers. Calculus isn't as rigorous as Real Analysis, so it doesn't matter anyway The focus is on computing the answers, not knowing how the maths works (which is what Real Analysis is responsible for).
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January 31st, 2016, 08:53 AM   #9
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Quote:
Originally Posted by Country Boy View Post
Yes, a variable can be an "independent variable" in one function, and a "dependent variable" in another. In the case where $\displaystyle y= f(x)$ and $\displaystyle x= f^{-1}(y)$, x is the "independent variable" in the first equation and the dependent variable in the second. y is the dependent variable in the first equation and the independent variable in the second. That is one reason why the terms "independent variable" and "dependent variable" are not very useful!
I respectfully disagree. They may not be 'meaningful' on a rigorous mathematical level, but these concepts have helped a lot when I learnt about functions. I've been brainwashed into thinking in terms of independent and dependent variables since I was 12, so it comes naturally for me (and probably the OP as well)...

For one thing, it helped me understand why something like y = x^2 isn't a function...

By the way, just noticed v8archie and Country Boy are now in the Maths Team. Congrats!
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January 31st, 2016, 10:35 AM   #10
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What do you mean by "something like y = x^2"? The equation y = x^2 would be an acceptable function definition if a complete formal definition doesn't need to be given.
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