I am stuck on this doubt :

Suppose $f=f(x,y,z).$ Hence, $ df= \frac {\partial f}{\partial x}dx + \frac { \partial f}{\partial y}dy + \frac {\partial f}{\partial z}dz.$

Then, is the following equation correct :

$$\frac {df}{dx}=\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}\frac{dy}{dx} + \frac {\partial f}{\partial z}\frac{dz}{dx} \,\,\,\,(*)$$

The reasoning used in obtaining $(*)$ is : "dividing" the whole equation by $dx$. Normally, the $\large \frac {d}{dx}$ operator is used in single variable calculus where only single-variable functions are differentiated wrt $x$. But it does look a bit awkward (at least to me) when used in multi-variable calculus. Do the expressions $\large \frac {df}{dx}$, $\large \frac {dy}{dx}$ and $\large \frac {dz}{dx}$ even make any sense when used like this ?

I know that "division" by $\partial x$ can cause problems in multi-variable calculus. But what about "division" by $dx$. It works fine in single-variable calculus.

If $\large \frac {df}{dx}$ makes any sense, then does it mean the "total" rate of change of $f$ wrt $x$ if $y$ and $z$ are allowed to change ?

Summary :

(1) Is division by $dx$ allowed in multi-variable calculus?

(2) What does $\frac {df(x,y,z)}{dx}$ mean if answer to (1) is "yes" ? Does it mean anything if the answer to $(1)$ is "no" ?

  • $\begingroup$ The notation $\frac{d}{dx}f(x,y,z)$ can make sense in the situation where $y$ and $z$ are given by functions in $x$, in other words, when you want to find the rate of change of $f\big(x,u(x),v(x)\big)$ w.r.t. $x$. But if $y$ and $z$ do not depend on $x$, the notation (*) makes little sense. And also, dividing by $dx$ (or $dy$, $dz$) is not a valid operation. You may argue that $\frac{d}{dx}F(x)$ has division by $dx$, but it is not really division by $dx$ even though the notation may suggest it. $\endgroup$ – WE Tutorial School Oct 6 '19 at 13:36

If $$f=f(x,y,z)$$ where $x,y,z$ are independent variables, then you may divide $df$ by $dx$.

For example $$f(x,y,z)= xyz+x^2+y^2+z^2$$

$$df = (yz+2x)dx + (xz+2y)dy + (xy+2z)dz$$

$$\frac {df}{dx} = yz+2x + (xz+2y)\frac {dy}{dx} + (xy+2z)\frac {dz}{dx} =yz+2x $$

Which is the same thing as $\frac {\partial{f}}{\partial {x}}$


Yes, it's correct, and what you've just computed is the rate of change of $f$ with respect to $x,$ where $y,z$ are also considered to be functions of $x.$


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