$u=xf(xy)$, show that $xu_{xx}-yu_{xy} = 0$

I need to show that:

$$xu_{xx}-yu_{xy} = 0$$

when $$u=xf(xy)$$

So, I did:

$$u_x = xyf_x(xy)+f(xy) \implies$$ $$u_{xx} = xy^2f_{xx}(xy)+2yf_x(xy)$$ and $$u_{xy} = xf_x(xy)+x^2yf_{xy}(xy)+xf_y(xy)$$

so:

$$xu_{xx} = x^2y^2f_{xx}(xy)+2xyf_x(xy)$$

$$yu_{xy} = x^2y^2f_{xy}(xy)+2xyf_y(xy)$$

but when I take one from another, I don't get $0$. This makes me think that there is a relation that says

$$f_{xx}(xy) = f_{xy}(xy)$$

so they both cancel, but I couldn't find it.

What am I doing wrong?

• your calculation of $u_x$ is wrong. Choose $u(x,y)=f(xy)=c$ then you get $u_x=c$ and that is wrong. Aug 16 '15 at 18:04
• How do you explain $u_x = xyf_x(xy)+f(xy)$? Aug 16 '15 at 18:06
• @miracle173 sorry, I wrote the wrong $u$ function. Please look now. Aug 16 '15 at 18:09
• Your calculation of $u_x$ is still wrong Aug 16 '15 at 18:09
• In fact it's not only wrong, it makes no sense! Which come to think of it may point the way to what you're missing. You talk about partial derivatives of $f$ here. The function $f$ does not have partial derivatives - it's a function of one variable! The notation $f(xy)$ means $f$ evaluated at $xy$. Not $f$ times $xy$. And not $f(x,y)$. Aug 16 '15 at 18:15

1 Answer

Note that $f$ is a sigle variable function, and it is composed with a 2-variable function, say, $w(x,y)=xy$. So $u$ is a 2-variable function.

We first calculate

• $u_x(x,y)=f(xy)+xyf'(xy)$.
• $u_{xx}(x,y)=yf'(xy)+y\left(f'(xy)+xyf''(xy)\right)=2yf'(xy)+xy^2f''(xy)$.
• $u_{xy}(x,y)=xf'(xy)+x\left(f'(xy)+xyf''(xy)\right)=2xf'(xy)+x^2yf''(xy)$.

Now: \begin{align*}xu_{xx}(x,y)-yu_{xy}(x,y)&=x\left(2yf'(xy)+xy^2f''(xy)\right)-y\left(2xf'(xy)+x^2yf''(xy)\right)\\ &=2xyf'(xy)+x^2y^2f''(xy)-2xyf'(xy)-x^2y^2f''(xy)\\ &=0.\end{align*}