If z is defined implicitly as a function of x and y, find $\frac{\partial z}{\partial x}$
$\begin{equation} \ yz = ln (x+z) \end{equation}$
I've attempted this equation going forward with implicit differentiation and I've used the theorem that states $ \frac{\partial z}{\partial x} = -\frac{\partial F/\partial x}{\partial F/\partial z}$
$\endgroup$2 Answers
$\begingroup$Given: $$\begin{equation} \ yz = \ln (x+z) \end{equation} \Rightarrow yz-\ln{(x+z)}=0 \Rightarrow F(x,y,z)=0,$$ we find: $$z_x=\frac{\partial z}{\partial x} = -\frac{\partial F/\partial x}{\partial F/\partial z}=-\frac{F_x}{F_z}=-\frac{-\frac{1}{x+z}}{y-\frac{1}{x+z}}=\frac{1}{xy+yz-1}.$$
$\endgroup$ $\begingroup$This should be:
Partially differentiating both sides with respect to x:
$y \ \frac{\partial z}{\partial x} = \frac{1}{x+z}(1 + \frac{\partial z}{\partial x})$
Now you can rearrange and obtain the correct value.
This way works because $z$ is an implicit function of $x$ and $y$.
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