In another thread, yourdadonapogos asked whether or not

 

[math] \frac{\partial}{\partial t} = \frac{d}{dt} [/math]

 

Dave didn't want to answer it in that thread, because that thread is devoted to ordinary derivatives, but I think the question is a good one. Does anyone know the answer?

 

Regards

well, that how the partial derivative defined. It's just an ordinary derivate where one of the variables is taken to be a constant.

well, that how the partial derivative defined. It's just an ordinary derivate where one of the variables is taken to be a constant.

 

I didn't exactly understand this.

 

Suppose you have the following function of x,y,z

 

[math] f(x,y,z) = 3xy^3 + 7zxy - 2x [/math]

 

The partial derivative of f with respect to x is given by

[math] \frac{\partial f}{\partial x} = 3y^3 + 7zy - 2 [/math]

 

During the differentiation process, the variables y,z were treated as constant.

 

So I do know that.

 

So they cannot be equivalent.

Well, they're obviously not equivalent except for one special case. For a function [math]f:\mathbb{R} \to \mathbb{R}[/math] we define the derivative by:

 

[math]\frac{df}{dx} = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}[/math]

 

For a function [math]f:\mathbb{R}^n \to \mathbb{R}[/math] we can define a whole host of partial derivatives for each component x_i:

 

[math]\frac{\partial f}{\partial x_1} = \lim_{h \to 0} \frac{f(x_1+h,x_2,\dots, x_n) - f(x_1,\dots,x_n)}{h}[/math]

 

[math]\frac{\partial f}{\partial x_2} = \lim_{h \to 0} \frac{f(x_1, x_2+h, x_3,\dots, x_n) - f(x_1,\dots,x_n)}{h}[/math]

 

And so on. The easy answer is that if n = 1, then the two definitions are equivalent.

yeah sorry, i should have made it more clear. Its quite easy to visualise partial derivatives when you have a function of two variables.

Thank you both.

 

PS: That notation was very good Dave.