2
$\begingroup$

Given $z=f(x,y)$, what's the difference between $\displaystyle\frac{dz}{dx}$ and $\displaystyle \frac{\partial f}{\partial x}$?

I got confused when I saw $dz/dx= \partial f/\partial x+\partial f/\partial y*dy/dx$, could somebody explain the difference to me? Thanks

$\endgroup$
1
  • 2
    $\begingroup$ Presumably $z(x) = f(x,y(x))$. $\endgroup$
    – copper.hat
    Nov 16, 2013 at 4:23

1 Answer 1

12
$\begingroup$

When you write $\displaystyle \frac{dz}{dx}$ you are assuming that $z$ is a function of $x$ only. This forces $y$ to be a function of $x$, therefore abusing the notation we write $y=y(x)$. Meanwhile, we have that $f$ continues to be a function of two variables, therefore $\displaystyle \frac{\partial f}{\partial x}$ means the usual partial derivative with respect to $x$.

When we take total derivatives we usually write it this way:

$$dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy.$$

If we assume $z$ is a function of a certain parameter $t$ then we'd have

$$\frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}.$$

In this case the parameter is $x$ itself, and we'd abuse notation by writing

$$\frac{dz}{dx} = \frac{\partial f}{\partial x} \frac{dx}{dx} + \frac{\partial f}{\partial y} \frac{dy}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}.$$

We are saying that "dx/dx = 1", essentially.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.