Differentiation of a Non-Function: is this possible?

[User1000]

I have a technical question that can simply be stated, "Is it possible to take the derivative of something that is not a function?"

I am aware that given an equation, one can still determine some form of piecewise solution or answer in terms of other variables that would provide an output for the slope at a given point on the original non-function; however, I'm not asking if this is possible. My question is simply, given that the definition of the derivative of f(x) is " f'(x)= lim(h->0) (f(x+h)-f(x))/h " is it technically possible to differentiate anything that is not a function (ie not passing vertical line test).

 

[Steven G]

I have a technical question that can simply be stated, "Is it possible to take the derivative of something that is not a function?"

I am aware that given an equation, one can still determine some form of piecewise solution or answer in terms of other variables that would provide an output for the slope at a given point on the original non-function; however, I'm not asking if this is possible. My question is simply, given that the definition of the derivative of f(x) is " f'(x)= lim(h->0) (f(x+h)-f(x))/h " is it technically possible to differentiate anything that is not a function (ie not passing vertical line test).

The answer to your question is yes. For example you can find the derivative wrt to t (or wrt to x etc) of x^2 + y^2 = pi/e^2 (any positive quantity)

 

[HallsofIvy]

Are you referring to a non-function relation such as \(\displaystyle x^2+ y^2= 1\)? You can always reduce a relation to several functions. For example, \(\displaystyle x^2+ y^2= 1\) can be written as \(\displaystyle y= \sqrt{1- x^2}\) and \(\displaystyle y= -\sqrt{1- x^2}\). From those, \(\displaystyle y'= \frac{x}{\sqrt{1- x^2}}\) for y non-negative and \(\displaystyle y'= -\frac{x}{\sqrt{1- x^2}}\) for y negative.

You can also do it using the "chain rule" in a manner called "implicit differentiation" differentiating each term with respect to x, \(\displaystyle 2x+ 2y(dy/dx)= 0\) so that \(\displaystyle dy/dx= -\frac{x}{y}\). Of course, with \(\displaystyle y= \sqrt{1- x^2}\) or \(\displaystyle y= -\sqrt{1- x^2}\), those are the same as before.

 

[Caesium]

I have a technical question that can simply be stated, "Is it possible to take the derivative of something that is not a function?"

I am aware that given an equation, one can still determine some form of piecewise solution or answer in terms of other variables that would provide an output for the slope at a given point on the original non-function; however, I'm not asking if this is possible. My question is simply, given that the definition of the derivative of f(x) is " f'(x)= lim(h->0) (f(x+h)-f(x))/h " is it technically possible to differentiate anything that is not a function (ie not passing vertical line test).

If your question is mathematical, I'm afraid it's ill-defined. What is "something that is not a function"? Is it a jeans pant, a gas station or magic of love?

On a related note, here's a function that cannot be even plotted, let alone be continuous, much less differentiated..