1

I have a question related to partial differential equations:

Say we have $f(x,y,g(x,z))$. Is $f_x\neq \frac{\partial f}{\partial x}$? By what I have read, $f_x)$ is the derivative of $f$ assuming $g(x,z)$ and $y$ to be constant.

  • They are supposed to mean the same thing, but sometimes notation can be abused and the meaning can become ambiguous. It depends on whether the "$x$" is supposed to represent the "first slot" or a parameter that may be scattered around various "slots". Your function can be thought of as $f(x_1,x_2,x_3)$ where $x_1=x$, $x_2=y$, and $x_3=g(x,y)$. These expressions could both mean $\partial f/\partial x_1$. – MPW Mar 01 '14 at 04:59

2 Answers2

0

They are the same whatever $f$ is. It is just a symbol to the first partial derivative of $f$ with respect to $x$. When you try to get this partial derivative, you have to assume that the independent variables be constant. In this case $g(x,y)$ depends on $x,y$, so you can not assume it as constant.

Semsem
  • 7,651
  • @Semsem- An additional question: "say we have the quasi-linear first order differential equation $a(x,y,u)u_x+b(x,y,u)u_y-c(x,y,u)=0$. This may have the implicit solution $u=u(x,y)$, which leads us to the equation $f(x,y,u)\equiv u(x,y)-u=0$." How does $u(x,y)-u$ give us ANY useful information, considering $u(x,y)-u=0$ for all $x,y$? Why can't we just write the solution in the form $f(x,y,u)=0$? –  Mar 01 '14 at 05:27
0

Personally, I would use $f_x=\frac{\partial f}{\partial x}$. However, perhaps the notation below would be helpful: $$ \frac{df}{dx} = D_1f+(D_3f)g_x $$ Or, if you like, view $x$ and $y$ as functions of $t$ so $$ \frac{d}{dt}(f(x,y,g(x,y)) = \frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dg}{dt} $$ But, setting $t=x$ this yields: $$ \frac{d}{dx}(f(x,y,g(x,y)) = \frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{dg}{dx} $$ and, as $\frac{dg}{dx} = \frac{\partial g}{\partial x}\frac{dx}{dx}+ \frac{\partial g}{\partial y}\frac{dy}{dx} = \frac{\partial g}{\partial x}$ $$ \frac{df}{dx} = \frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial x} $$ Moral of story: don't use the coordinate as a parameter unless you are prepared to deal with this ambiguity.

James S. Cook
  • 16,755