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Say I have variables $x,y_1,y_2,z_1,z_2$ all $\in \mathbb{R}$

And I have the following equations:

$$x = f_1(y_1,y_2)$$ $$y_1 = f_2(z_1,z_2)$$

How does:

$$dx \over dz_1$$

differ from:

$$\partial x \over \partial z_1$$

or am I confused?

Intuitively I just want to think about how $x$ varies in proportion to an infinitesimally small perturbation of $z_1$, so I don't understand the difference between the two different notations (nonpartial vs partial)?

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As a note, I would avoid the use of the word "normal" in your title. "Normal derivative" might be interpreted by some to mean the gradient in the normal direction! –  Arkamis Oct 26 '12 at 19:51
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Let's say you had $x=z_1z_2$. Then $\frac{\partial x}{\partial z_1} = z_2$ but $\frac{dx}{dz_1} = z_2 + z_1\frac{dz_2}{dz_1}$. (I haven't put this as an answer because you probably need more explanation than this.) –  Clive Newstead Oct 26 '12 at 19:56
    
Now I am really confused. If we plot in 3D a surface $x=z_1z_2$, then at a given point on the surface the tangent in the $z_1$ direction will correspond to the partial derivative $\partial x \over \partial z_1$. Does the total derivative have a similiar geometric interpretation? –  Andrew Tomazos Oct 26 '12 at 20:06
    
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The total derivative corresponds to the tangent plane to the surface at the given point; of course this plane is spanned by the tangent lines corresponding to $\partial x/\partial z_1$ and $\partial x/\partial x_2$. In higher dimensions the total derivative corresponds to the tangent space (whose dimension $n$ is the same as that of the given manifold), which is spanned by the $n$ first partials. –  Tabes Bridges Oct 26 '12 at 20:32

1 Answer 1

When a function is defined on more than one variables we use $\cfrac{ \partial f }{\partial x}$ to denote the partial derivation of $f$ with respect to one of its variables $x$ while holding the other variables constant.

This is a question of notation. Using $\cfrac {df} {dx}$, the total derivative, will show anyone that sees it that $f$ is only defined on the variable $x$ or that the other variables of $f$ are functions also defined on $x$.

  • Say for example a function $f(x_1, x_2, x_3)$ such that $x_1$, $x_2$, $x_3$ are independent. Then $$\cfrac {\partial f}{\partial x_1} =\left ( \cfrac {\partial f}{\partial x_1}\right )_{x_2, x_3}$$ is the partial derivative of $f$ with respect to $x_1$ holding $x_2, \space x_3$ constant.

  • But consider a function $f(t,x_1(t), x_2(t), x_3(t))$ then
    $$ \cfrac {df}{dt} = \cfrac {\partial f}{\partial t}\cfrac {dt}{dt}+\cfrac {\partial f}{\partial t}\cfrac {dx_1}{dt}+\cfrac {\partial f}{{\partial t}}\cfrac {dx_2}{dt}+\cfrac {\partial f}{{\partial t}}\cfrac {dx_3}{dt}$$ is the total derivative of $f$ with respect to $t$.

  • Now let's compute the total derivative of the first function $f(x_1, x_2, x_3)$ with respect to $x_1$ $$\cfrac {df}{dx_1} =\cfrac {\partial f}{\partial x_1}\cfrac {dx_1}{dx_1}+\cfrac {\partial f}{\partial x_1}\cfrac {dx_2}{dx_1}+\cfrac {\partial f}{{\partial x_1}}\cfrac {dx_3}{dx_1} = \cfrac {\partial f}{\partial x_1}$$ which is basically the same thing but only because the variables are independent of one another. It changes if one of them say $x_3$ is a function of $x_1$ then $\cfrac {dx_3}{dx_1} \ne 0$ and then $$\cfrac {df}{dx_1} = \cfrac {\partial f}{\partial x_1} + \cfrac {\partial f}{{\partial x_1}}\cfrac {dx_3}{dx_1} $$

See Partial Derivatives and Total Derivatives.

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So what does $dx \over dz_1$ mean? Or is it undefined? If it is undefined why not just use the same notation? –  Andrew Tomazos Oct 26 '12 at 19:43
    
It is defined only if the other variables are constants. If the other variables are indeed variable then one has to use the right notation. –  user31280 Oct 26 '12 at 19:46
    
But for all mechanical purposes all the same rules apply in both cases? We are asking how the "numerator" variable varies with respect to the "denomiator" variable, assuming everything else is held constant. –  Andrew Tomazos Oct 26 '12 at 19:54
    
I got it all wrong, $dx \over dz_1 $, the total derivative implies that every other variable is a function of $z_1$. –  user31280 Oct 26 '12 at 19:58
    
So what is here called $\frac{df}{dt}$ is really just $g'(t)$ where $g(t) = f(t,x_1(t),x_2(t),x_3(t))$. I know that sometimes $\frac{df}{dt}$ is used in this situation, but to me it seems like sloppy notation. –  littleO Oct 26 '12 at 20:29

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