What constitutes a derivative?

[hobz]

Suppose I have a small (infinitesimal) quantity [math]dy[/math] and another small quantity [math]dx[/math] and they are related by [math]dy = k \cdot dx[/math]. Will that automatically imply that [math]\frac{dy}{dx}=k[/math] is the derivative [math]\left(\frac{\mathrm{d}y}{\mathrm{d}x}=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)- f(x)}{\Delta x}\right)[/math] of [math]x[/math] with respect to [math]y[/math]?

 

I have seen several examples of such things occuring in engineering textbooks, such as electrical relations between the charge on a capacitor and the voltage across it. (I can't remember the details, and my notes are safely stored in the basement).

[Mr Skeptic]

Yes. You have y(x) = kx, and then [math]dy = \frac{\partial y}{\partial x} dx = \frac{\partial (kx)}{\partial x} dx = k dx [/math]

[triclino]

Suppose I have a small (infinitesimal) quantity [math]dy[/math] and another small quantity [math]dx[/math] and they are related by [math]dy = k \cdot dx[/math]. Will that automatically imply that [math]\frac{dy}{dx}=k[/math] is the derivative [math]\left(\frac{\mathrm{d}y}{\mathrm{d}x}=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)- f(x)}{\Delta x}\right)[/math] of [math]x[/math] with respect to [math]y[/math]?

 

I have seen several examples of such things occuring in engineering textbooks, such as electrical relations between the charge on a capacitor and the voltage across it. (I can't remember the details, and my notes are safely stored in the basement).

 

Given y = f(x) ,then we define : dy = f'(x)Δx and Δx= 1.dx .

 

Hence by definition: dy/dx = f'(x) ,where dy/dx is the ratio of the differentials dy ,dx

 

In the case where the derivative is denoted by :[math]\frac{dy}{dx}[/math],then this is equal with the ratio of the two differentials dy/dx