Suppose we have a function $f(x,y) = 2x^2+2y^2$

Next we change the variables with $u(x,y)=2x^2+2y^2$ so we have $f(x,y)=g(u(x,y))=u(x,y)$

Since these functions are equal everywhere, we can write $f(x,y)=g(u(x,y))$ and take partials of both sides.

Question: What is the formula for $\frac{\partial}{\partial y}\frac{\partial}{\partial x} g(u(x,y))$ (in general, not for the specific above example)

For example, we can say that $$ \tag{1} \frac{\partial}{\partial x} f(x,y) = \frac{\partial}{\partial x} g(u(x,y)) = \frac{\partial g }{\partial u} \frac{\partial u }{\partial x} $$

But I want such a a relationship for cross partials.

(note, both the LHS and RHS of (1) both are $4x$ for the example I have given)

Note: I am using $\frac{\partial}{\partial y}\frac{\partial}{\partial x}$ to mean first take the partial w.r.t to $x$ and then the partial of that partial w.r.t $y$. I may have the order backwards; I always forget what is the correct order to write the partials in.


1 Answer 1


Sincce $f_x=g^\prime(u)u_x$, $f_{xy}=g^{\prime\prime}(u)u_xu_y+g^\prime(u)u_{xy}$.

  • $\begingroup$ Ahhh, thank you. I was trying to do it in the notation with fractions used in my post and was getting confused. $\endgroup$
    – user106860
    Commented Nov 13, 2020 at 15:30

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