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I have been beating my head against this question for quite some time, I do not know whether it has been asked before, but I can't find any information about it!

I am taking Calculus 1 course and I cannot grasp the concept of a derivative. From what I understand, a derivative is a function with the following signature:

$$(\text{derivative with respect to particular free variable}) :: (\lambda x \to (f)\; x) \to (\lambda x \to (f') x)$$

also phrased as

$$(\text{derivative with respect to particular free variable}) = ((\lambda x \to (f) x) \to (\lambda x \to (f') x))$$

e.g:

$$(\text{derivative with respect to x}) (\lambda x \to x^2) = (\lambda x\to 2\cdot x)$$

This makes sense but one thing bothers me: what does "derivative with respect to x" mean? In particular, in single variable Calculus this notation assumes $x$ is always a particular variable such as ['a'..'z'].

This works fine for basic derivatives such as: $$(\text{derivative with respect to x}) (\lambda x \to \ln x) = (\lambda x \to \tfrac{1}{x})$$

What I would like to understand is: Why does (derivative with respect to x) make sense but $(\text{derivative with respect to} (\lambda x \to 2))$ and

$(\text{derivative with respect to} (\lambda x \to \ln(x)) $ do not seem to make any sense to me.

in classical terms, I cannot do (derivative of $\ln x$ with respect to $1$) nor (derivative of $\ln x$ with respect to $\ln x$) without my head starting to hurt, because those concepts were not taught to me yet, or I have not payed enough attention to understand them.

Can somebody please explain what the following two really mean?

  1. Derivative of $f(x)$ with respect to a constant such as $1,2,3,\ldots 9999$
  2. Derivative of $f(x)$ with respect to a function such as $\ln(x)$, $\sin(x)$, $\cos(x)$

Thanks ahead of time, this has been bothering me for quite a few years!

$\langle$Editor's note: I've left the following in the post for archival's sake.$\rangle$

PS: I am terrible at formatting so to the great ones responsible for formatting noob's questions (I thank you much for your work)

  1. convert \ to lambdas
  2. convert d/dx to symbolic d/dx notation (not the worded derivative ones)
  3. convert arrows to arrows used in set theory/category theory
  4. keep the "(derivative of ... with respect to ...)" as they are, as I have no idea how to express them differently, dA/dB doesn't seem to make sense to me since derivatives are taught to be polymorphic function rather than a function of two variables, and division only makes it even more confusing due to the abuse of notation. (Feel free to give me a link to study formatting, I can't find it).
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  • $\begingroup$ So it always has to be a free variable, and the function MUST be expressed through the free variable (which can be a function which must also be free)? (Response to apparently deleted post). $\endgroup$ – Dmitry Apr 13 '13 at 22:51
  • $\begingroup$ I've edited your question to use $\LaTeX$ (the math formatting we use). Please make sure the post still represents your original intent. A great resource for learning formatting here is this meta question. $\endgroup$ – apnorton Apr 13 '13 at 23:18
  • $\begingroup$ We do not differentiate expressions. We differentiate functions. The notation $d/dx (\mbox{expression in }x)$ simply means "define a function using the given expression, differentiate that function, and evaluate at $x$". As for differentiating "with respect to" something other than a free variable - I've never seen anyone try to define or use this. $\endgroup$ – wj32 Apr 13 '13 at 23:37
  • $\begingroup$ I think that notion is confusing because the result of the d/dx is intended to be the function of the slope at any given point of the given function, howerver (d/dx)x is a constant(1 in this case), which is not a function, and performing function application of a constant on anything is illogical(eg, lisp would yell at you for trying to do (1 2)), so I am trying to distance myself from this definition. The alternative defines (d/dx)f = (\x -> (x -> f'x)) which when applied onto x, provides a function (\x -> (x -> 1)), which causes no contradictions. $\endgroup$ – Dmitry Apr 13 '13 at 23:43
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Derivatives are usually defined in terms of limits. The derivative of $f(x)$ with respect to $g(x)$ can be defined as $$\lim_{h\to0}{f(x+h)-f(x)\over g(x+h)-g(x)}$$ provided the limit exists. In the case $g(x)=x$, this reduces to the familiar formula for the derivative of $f(x)$ with respect to $x$, $$\lim_{h\to0}{f(x+h)-f(x)\over h}$$ In the case where $g(x)$ is a constant, the denominator $g(x+h)-g(x)$ is identically zero, so the limit n'existe pas. This could explain why no one ever differentiates with respect to a constant.

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  • $\begingroup$ Thanks that solved most of my question! So both top and bottom must approach 0, and you want to determine their rate of one with respect to another. So by the nature of derivative, you only care about limits of something/h or g/h as g and h approach 0. constant/h would be undefined, g/h could be defined. Would this mean that derivative of constant with respect to constant and function with respect to constant is nothing more than a finite difference? $\endgroup$ – Dmitry Apr 13 '13 at 23:51
  • $\begingroup$ Also, does that mean that h->0 is essentially synonymous to dx in this case? Or do I misunderstand. Seems wrong. $\endgroup$ – Dmitry Apr 14 '13 at 0:02
  • $\begingroup$ Derivative of constant w.r.t. constant would lead to $\lim_{h\to0}(0/0)$, which is undefined. Derivative of function w.r.t. constant, as shown above, does not exist. I don't see how this can be interpreted as a finite difference. And I don't know what to make of "does that mean that $h\to0$ is essentially synonymous to $dx$". $\endgroup$ – Gerry Myerson Apr 14 '13 at 0:10
  • $\begingroup$ One more thing, does dsin(x) and sin(x)dx make any sense by themselves, without the respective denominator dx and integral? $\endgroup$ – Dmitry Apr 14 '13 at 0:41
  • $\begingroup$ They make sense, if you define them in a sensible way. You could, for example, define $df(x)$ to mean $f'(x)\,dx$ and, under certain circumstances, you might find it useful to do so. If I were teaching an intro Calculus course, though, I'd probably steer clear of such notations --- students have enough trouble with the usual ones, they don't need more. $\endgroup$ – Gerry Myerson Apr 14 '13 at 6:27
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dy/dx in words means change in y with one unit small change in x so in cases where x is a constant that simply means there is no change in x against which you would otherwise look at change in y so simply tht implies dx=0 and hence dy/dx is not defined.

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