If I have a certain function $f(x,y,z)$ where $x$ is an independent variable constrained such that $g(x,y)=c$ then when I calculate $\partial f/\partial x$ what happens to $z$ in the solution?
Say for example my $w$($x$)= $z$$x$$e^y$+ $x$$e^z$+ $y$$e^z$ and $g$($x$)=$x$$y^2$+$x^2$$y$=1 and I want to calculate $\partial w/\partial x$ assuming $x$ is an independent variable.
So I first find the total differential of $w$ that goes like $d$$w$ = ($z$$e^y$+$e^z$)$d$$x$ + ($z$$x$$e^y$+$e^z$)$d$$y$ + ($x$$e^y$+$x$$e^z$+$y$$e^z$)$d$$z$
and then find $d$$y$ in terms of $d$$x$ using $g$($x$,$y$) that is by differentiating g(x,y)= $x$$y^2$+$x^2$$y$=1
$d$$g$=($2$$x$$y$+$y^2$)$d$$x$+($2$$x$$y$+$x^2$)$d$$y$=0
which gives $d$$y$=-($2$$x$$y$+$y^2$)/($2$$x$$y$+$x^2$)$d$$x$ and take $d$$z$=0 since $g$($x$,$y$) does not have $z$ and subsitute it in $d$$w$
$d$$w$ = ($z$$e^y$+$e^z$)$d$$x$ - ($z$$x$$e^y$+$e^z$)($2$$x$$y$+$y^2$)/($2$$x$$y$+$x^2$)$d$$x$
which gives
$d$$w$ = [($z$$e^y$+$e^z$)($2$$x$$y$+$x^2$)-($z$$x$$e^y$+$e^z$)($2$$x$$y$+$y^2$)/($2$$x$$y$+$x^2$)] $d$$x$
but the solution provided to me somehow does not match with my results. I do not know if my understanding is wrong or the solution is incorrect
Using the Implicit function theorem, you get from $g(x, y(x)) = c$:
$\frac{dg}{dx} = 0 = \frac{\partial g}{\partial x} + \frac{\partial g}{\partial y} \frac{dy}{dx}$, that is if $\frac{\partial g}{\partial y}$ is invertible:
$\frac{dy}{dx} = - \left(\frac{\partial g}{\partial y} \right)^{-1} \frac{\partial g}{\partial x}$.
We can know differentiate $f(x,y(x),z)$:
$\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx} = \frac{\partial f}{\partial x} - \frac{\partial f}{\partial y} \left(\frac{\partial g}{\partial y} \right)^{-1} \frac{\partial g}{\partial x}$