0
$\begingroup$

$x,y$ are independent variables and $f = f(x,y)$. Some other variable $z = z(x,y)$. I want to calculate $\frac{df}{dz}$.

I started as follows, $$\frac{df}{dz} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial z} $$ Is it correct?

Suppose $f = x+y$ and also $z=x+y$, so $\frac{df}{dz} = 1$. On the other hand $\frac{\partial f}{\partial y} = 1=\frac{\partial f}{\partial x} $ and $\frac{\partial y}{\partial z} =1 = \frac{\partial x}{\partial z}$

so, $\frac{\partial f}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial z} = 2$

can anyone help me figure out what went wrong?

$\endgroup$
  • 1
    $\begingroup$ ${\partial x\over \partial z} \ne \left({\partial z\over\partial x}\right)^{-1}$. You can’t simply take the reciprocals of partial derivatives to get partial derivatives of the inverse function as you can with ordinary derivatives of single-variable functions. $\endgroup$ – amd Jun 6 at 9:15
1
$\begingroup$

What do you mean/hope to calculate by $\mathrm df/\mathrm dz$? It would be relatively common to write $\mathrm df=\dfrac{\partial f}{\partial x}\,\mathrm dx+\dfrac{\partial f}{\partial y}\,\mathrm dy$ and similarly for $z$ in place of $f$. But then if you tried to write $\mathrm df/\mathrm dz$, you'd usually get weird things that don't simplify (or even make sense?) like $\dfrac{5\,\mathrm dx+x\,\mathrm dy}{xy^2\,\mathrm dx+e^y\,\mathrm dy}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.