Compute $\dfrac{\partial f(x,y,g(x,y))}{\partial x}$ Question is in the title... I was wondering if someone could help me with that partial derivative, preferably with the total derivative if possible. Thanks a lot
 A: The chain rule gives us a general formula
$$\frac{\partial}{\partial x} f(\alpha,\beta,\gamma)=\frac{\partial f}{\partial\alpha}\frac{\partial\alpha}{\partial x}+\frac{\partial f}{\partial\beta}\frac{\partial\beta}{\partial x}+\frac{\partial f}{\partial\gamma}\frac{\partial\gamma}{\partial x}. $$
Now what if $\alpha=x$ and $\beta=y$ and $\gamma=g(x,y)$?

To summarize the discussion below: partially differentiating $f$ with assumed independent arguments, and then evaluating at dependent arguments, obtains different results from first creating a new function by evaluating $f$ at dependent arguments and then differentiating that.
A: To make things clearer, I thought I'd write a comment, but this is too big. Let $h(x,y) = f(x,y,g(x,y))$. What I believe that you want to compute is $\frac {\partial h}{\partial x}$, which is different from $\frac{\partial f}{\partial x} f(x,y,z)$ where $z$ happens to be such that $z = g(x,y)$. In other words, you're taking the limit
$$
\lim_{t \to x} \frac{f(t,y,g(\overset{\downarrow}{t},y)) - f(x,y,g(x,y))}{t-x} = \frac{\partial h(x,y)}{\partial x}
$$ 
which is not the same thing as taking the limit
$$
\lim_{t \to x} \frac{f(t,y,g(\overset{\downarrow}{x},y)) - f(x,y,g(x,y))}{t-x} = \left. \frac{\partial f(x,y,z)}{\partial x} \right|_{z = g(x,y)}
$$
(notice how in one expression the $x$ in $g(x,y)$ varies as $t$, but in the second one the $x$ remains fixed as $t$ varies). The first expression measures the variation of $f(x,y,g(x,y))$ when $x$ varies, but the second expression measures the variation of $f(x,y,z)$ when $x$ varies, then evaluates the derivative at $z = g(x,y)$. Those two are totally different.
Hope that helps, feel free to comment.
