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What is meant by "take the derivative of a function"?

  1. Sep 28, 2015 #1
    Bourbaki defines a function as follows: We give the name of function to the operation which associates with every element x the element y which is in the given relation with x; y is said to be the value of the function at the element x, and the function is said to be determined by the given functional relation. Two equivalent functional relations determine the same function.

    This means that f(x), an analytic expression for example, is the image of x under f, and not the function itself, just as y, the output value, is not the function itself. Thus, why do we say "take the derivative of f(x)" when f(x) is not even the function, but rather the image of the variable x under f? In addition, the notation dy/dx gives the feeling that the input to a differentiation operator is not a function at all, but an expression that defines a function, y=f(x). Given that we say "take the derivative of such and such function", shouldn't we always write D(x↦x2)=(x↦2x) rather than D(x2)=2x. Is the latter just a shorthand for the more rigorous former?
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  3. Sep 28, 2015 #2


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    Yes, it's just a shorthand. In the vast majority of cases the meaning is perfectly clear and causes no problems. But in some circumstances it can cause ambiguity, and then the author needs to take more care.
  4. Sep 28, 2015 #3
    Why don't we say that a differentiation operator takes an analytic expression (such as a combination of elementary functions) as input and returns and analytic expression as output, since that's really what the differentiation operator does? We see ##\displaystyle \frac{d}{dx}x^2 = 2x## in real calculations rather than ##\displaystyle \frac{d}{dx}(x \mapsto x^2) = (x \mapsto 2x)##
  5. Sep 28, 2015 #4


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    Not all differentiable functions are analytic.

    How would you write diff eqns under such an approach?
  6. Sep 28, 2015 #5
    I guess you're right. But here's what I understand: something like y = 5x or y = ln(x) describes a condition for a function, just as a table or a graph could describe a condition. y is not the function nor is 5x or ln(x). This is why I don't get why we say a derivative takes in functions and outputs functions. y is not a function, yet that is what we input into d/dx. Thus it would seem as though derivatives should be defined such that they take in expressions that define a function and output expressions that define another function. You can't calculate a derivative without an expression...
  7. Sep 28, 2015 #6


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    Correct. We assign the value of the function to the variable y.
    Yes. Given a function f(x), the derivative (with respect to x) is [itex] \frac{df(x)}{dx}[/itex].
  8. Sep 29, 2015 #7
    If y is not the function but rather the value of the function, then what does ##\frac{dy}{dx}## really mean? Why would this be the correct type of argument in the d/dx operator?
  9. Sep 29, 2015 #8


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    You can think of dy/dx as the rate of change of y with respect to change in x. As andrewkirk said in an earlier post, it's just shorthand.
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