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Let $f:\Omega\subset\mathbb{R^2\to\mathbb{R}}$ be a function such that $f\in\mathit{C^1}(\Omega)$. Now, consider the function:

$$g(x,y,z):=x^4f(y/x,z/x)$$

Prove that

$$x\frac{\partial g}{\partial x}+y\frac{\partial g}{\partial y}+z\frac{\partial g}{\partial z} = 4g$$

Now, I'm not sure whether I'm calculating those partial derivatives properly, but I conclude that:

$$x\frac{\partial g}{\partial x}=4x^4f(y/x,z/x)+x\cdot x^4\frac{\partial f}{\partial x}$$ $$y\frac{\partial g}{\partial y}=y\cdot x^4\frac{\partial f}{\partial y}$$ $$z\frac{\partial g}{\partial z}=z\cdot x^4\frac{\partial f}{\partial z}$$

Now, note that $x\frac{\partial g}{\partial x}=4g+x^5\frac{\partial f}{\partial x}$, which means its enough to prove that:

$$x\cdot x^4\frac{\partial f}{\partial x}+y\cdot x^4\frac{\partial f}{\partial y}+z\cdot x^4\frac{\partial f}{\partial z}=0$$

This is where I'm stuck, I have no clue how to proceed. I know that, since $f\in\mathit{C^1}(\Omega)$, it is differentiable. Not sure how to use this though. I'm expecting maybe all 3 partial derivatives are $0$, which would prove the statement. Or maybe they just compensate each other. ¿Any ideas?

F.Webber
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1 Answers1

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Careful with your partial derivatives. Writing subscripts for them, $$g_x = 4x^3 f(y/x,z/x) + x^4 \left(-y/x^2 f_1(y/x, z/x) - z/x^2 f_2(y/x, z/x)\right)$$

by the chain rule where $f_1$ and $f_2$ are the derivatives of $f$ with respect to the first and second variable. Hence

$$g_x = 4x^3f(y/x,z/x) -yx^2f_1(y/x, z/x) - zx^2f_2(y/x, z/x)$$

Next, $$g_y = (1/x) \cdot x^4f_1(y/x,z/x) + 0 \cdot x^4 f_2(y/x,z/x) = x^3f_1(y/x,z/x)$$ and similarly, $$g_z = 0 \cdot x^4 f_1(y/x,z/x) + (1/x)\cdot x^4f_2(y/x,z/x) = x^3f_2(y/x,z/x)$$

Hence

$$xg_x + yg_y + zg_z = 4x^4f(y/x,z/x) = 4g $$

Simon S
  • 26,524