# 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

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Are $x$ and $y$ independent? –  anon Mar 13 '12 at 5:24
Yessir [char limit] –  ae0709 Mar 13 '12 at 5:25
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## 2 Answers

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.

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so $f_x(x,y,g(x,y))=f_x\cdot1 + f_x\cdot0 + f_x\cdot g_x(x,y)$? I don't think that's right because the left-hand side appears as a term in the right hand side. –  ae0709 Mar 13 '12 at 5:28
@aengle: You're confused. The partial derivative of $f(x,y,g)$ wrt $x$ is not the same as $f_x$ evaluated at $(x,y,g)$ - which is what you have written in your comment. Furthermore the RHS of your equation should not have a subscript $x$ on each individual instance of $f$; look at the chain rule more closely! It should be $\frac{\partial}{\partial x}\big[f(x,y,g(x,y))\big]=f_x\cdot1+f_y\cdot0+f_g\cdot\frac{\partial g}{\partial x}.$ The term $f_g$ should be understood to mean "the derivative of $f$ wrt its third argument, evaluated at $(x,y,g(x,y))$." –  anon Mar 13 '12 at 5:30
Ok, so $\frac{\partial}{\partial x} f(x,y,g)=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}+\frac{\partial f}{\partial g(x,y)}\frac{\partial g(x,y)}{\partial x}$. So what is the difference between the term on the left and the first on the right in this case? (and thanks for the help :) ) –  ae0709 Mar 13 '12 at 5:39
@aengle: Consider a simpler case, $f(u,v)=u+v$. Now $f_u=1$. But if we let $v=v(u)=u^2$ be dependent on $u$, then $\frac{\partial}{\partial u}f(u,v(u))=\partial(u+u^2)/\partial u=1+2u$. If we simply evaluated $f_u$ at $(t,t^2)$ we'd still get $1$. Why are these different? Because in the former case you formally treat all arguments as independent to create a partial derivative of $f$, while in the latter case you're really partially differentiating a function created from $f$ - created by evaluating $f$ at dependent arguments before any differentiation. Obviously the results may differ. –  anon Mar 13 '12 at 5:47
OK. That maeks sense. Thanks a lot! –  ae0709 Mar 13 '12 at 5:51
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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.

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I understand what you mean. This wasn't my question at all actually, but it really does help clarify the subtle difference that I wasn't 100% understanding until after reading this. Thank you :) –  ae0709 Mar 13 '12 at 23:01
It wasn't your question, but it was your problem. =P –  Patrick Da Silva Mar 13 '12 at 23:04
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