chetzread said:
well , i couldn't understand why the author wrote ∂f/∂x + (∂f/∂z)(∂z/∂x) = 0 , i have no idea how he come up with this equation...In the notes, he said that by applying chain rule to LHS of equation? which equation did he mean?
We'll humour mfb and write$$f(x,y,z) = f\left (x,y,z\left (x,y\right )\right ) = 0$$ Apply the definition of derivative $$ {dg\over dx} \equiv\ \ \lim\limits_{h\downarrow 0} {g(x+h)-g(x)\over h}$$while y, as the other independent variable, is kept constant. Since ##f=0## and stays 0, this derivative should give 0.
this is how the author was lured into writing ##\displaystyle {\partial f\over\partial x}##
and I don't know how to do it didactically foolproof either. The point is that y is kept constant (because it is an independent variable) but z is a function of x so z does vary -- and thereby influences the derivative of f wrt x.
So what we get is $$ 0 =
\ \ \lim\limits_{h\downarrow 0} {f(x+h, y, z(x+h,y)) - f(x,y,z(x,y))\over h} = \\
\quad \lim\limits_{h\downarrow 0} {f(x+h, y, z(x+h,y)) - f(x,y,z(x+h,y) )\over h} + {f(x, y, z(x+h,y)) - f(x,y,z(x,y) )\over h}
$$
First term is ##\displaystyle {\partial f\over\partial x}## a.k.a. ##f_x##
Second term is ##\displaystyle {\partial f\over\partial z} \displaystyle {\partial z\over\partial x}## a.k.a. ##f_z z_x##
In summary: $$f_x + f_z z_x = 0$$
Look up chain rule in your textbook.
