hobz Posted April 15, 2010 Share Posted April 15, 2010 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). Link to comment Share on other sites More sharing options...

Mr Skeptic Posted April 16, 2010 Share Posted April 16, 2010 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] Link to comment Share on other sites More sharing options...

triclino Posted April 16, 2010 Share Posted April 16, 2010 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 Link to comment Share on other sites More sharing options...

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