I want to calculate $\cfrac{\partial^2 }{\partial x\partial y}f(x^2-y^2,xy)$ in terms of partial derivatives of the function $f$, assuming that second-order partials of $f$ are continuous.

I move on as follows:

Set $f=(u(x,y),v(x,y))$. We write $f_1$ for partial derivative of $f$ w.r.t. first component so $\cfrac{\partial f}{\partial u} = f_1$ and similarly $\cfrac{\partial f}{\partial v} = f_2$.

So, $\cfrac{\partial f}{\partial y} = \cfrac{\partial f}{\partial u} \cfrac{\partial u}{\partial y} + \cfrac{\partial f}{\partial v}\cfrac{\partial v}{\partial y} = f_1(u,v)(-2y) + f_2(u,v)x $.

Now, we have to find $\cfrac{\partial }{\partial x} \cfrac{\partial f}{\partial y} $ and that is the part that I do not know how to do it. I am doing as this way:

$\cfrac{\partial }{\partial x} \cfrac{\partial f}{\partial y} = \cfrac{\partial }{\partial x}\big[f_1(u,v)(-2y) + f_2(u,v)x \big] = \cfrac{\partial }{\partial x}\big(f_1(u,v)(-2y)) + \cfrac{\partial}{\partial x}(f_2(u,v)x)$ $ (*)$.

$=[-2y( \cfrac{\partial f_1}{\partial u} \cfrac{\partial u}{\partial x} + \cfrac{\partial f_1}{\partial v}\cfrac{\partial v}{\partial x})]+[x.(\cfrac{\partial f_2}{\partial u} \cfrac{\partial u}{\partial x} + \cfrac{\partial f_2}{\partial v}\cfrac{\partial v}{\partial x})]$

$=-2y.(f_{11}(u,v)2x + f_{12}(u,v)y) +1.(f_{21}(u,v)2x +f_{22}(u,v)y)$

However, the book says that we have to use product rule probably at $(*)$. Can you explain how to solve this problem explicitly? Thank you.

  • 1
    $\begingroup$ Well, yes, $f_2(u. v)x$ is a product. It is the product of $f_2(u, v)$ and $x$. Its derivative with respect to x is the part you give $\frac{\partial f_2}{\partial u}\frac{\partial u}{\partial x}$ but then you need to add $f_2(u. v)\frac{dx}{dx} = f_2(u, v)$. Strictly speaking the same is true of $\frac{\partial }{\partial x}(f_1(u.v)(-2y))$ but. of course, the derivative of -2y with respect to x is 0. $\endgroup$ – user247327 Oct 19 '17 at 15:33

Yes, the Product Rule has to be applied at the (*) step. Look at the second term in (*): $$\frac{\partial}{\partial x}\left(f_2(u,v)x\right)=\frac{\partial}{\partial x}\left(f_2(u,v)\cdot x\right).$$ This is the derivative with respect to $x$ of a function that is a product of $f_2(u,v)=f_2(u(x,y),v(x,y))$ and $x$. Since both parts of this product depend on $x$, you have to apply the Product Rule to find its derivative correctly.

Your mistake is that you took out $x$ outside of this derivative as if it's a constant. But the derivative here is with respect to $x$, so $x$ is NOT constant here.


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