I know how to apply the generalized chain rule for functions of this type:
$z = u^2 + v^2$, $u = x^2y$ and $v = xy$

$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} +\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$ etc.

I came across a function of this type:
$z = f(u, v)$, $u = x^2y^2$ and $v = 5x + 1$

Because of the $f(u,v)$, I don't know how to proceed! Anyone could give me a hint on how to find partial derivatives of this function? What is it that I don't understand?


  • 3
    $\begingroup$ Uh, didn't you already ask this question? math.stackexchange.com/questions/308504/… $\endgroup$
    – wj32
    Feb 23, 2013 at 3:20
  • $\begingroup$ I think you're not understanding the notation for partial derivatives ("$\partial$"). <rant> I don't blame you. It's vague and confusing. If you really want to understand what's going on, you have to start thinking "Here's a function $z:\mathbb{R}^2\to\mathbb{R}$ that is a composition of functions. What is its derivative?" rather than "Here's some expression $z=f(u,v)$. What is $\partial z/\partial x$?", which makes no sense. $\endgroup$
    – wj32
    Feb 23, 2013 at 3:53
  • $\begingroup$ ... On the LHS of your "chain rule" you write $\partial z/\partial x$ and on the RHS you write $\partial z/\partial u$. So $z$ has magically changed from a function of $x$ to a function of $u$ and $v$. The same letter is being used to denote two different things. This kind of notation makes the chain rule hard to understand rigorously. </rant> In this case, $z=f\circ g$ where $g(x,y)=(x^2y^2,5x+1)$. $\endgroup$
    – wj32
    Feb 23, 2013 at 3:54

1 Answer 1


If there is no other information on $f$, just keep it as is and proceed. Of course, you'll end up with terms containing $\frac{\partial f}{\partial u}$ and others, $\frac{\partial f}{\partial v}$.

The question does not seem to imply any need to go further. Just obtain the "general form" of the answer.


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