Show that $\frac{\partial^3u}{\partial x\partial y\partial z} = F(xyz)$

Question

$u=f(xyz)$. Show that $\frac{\partial^3u}{\partial x\partial y\partial z} = F(xyz)$ and find $F$. I have no idea what should I do?

Answer 1

Hint. Let $v=xyz.$ Then begin, using the chain rule to have that $$\partial u/\partial z=F'_vv'_z.$$

Answer 2

If $f$ is a function from $\mathbb{R}\rightarrow \mathbb{R}$ which is three times differentiable, and $f',f'',f'''$ its first three derivatives, we have

$$\frac{\partial^3}{\partial x \partial y \partial z}f(xyz) = \frac{\partial^2}{\partial y \partial z} yz f'(xyz)$$ by performing the partial derivative with respect to $x$ and using the chain rule $\frac{\partial}{\partial x} f(xyz) = f'(xyz)$.

Now $$ \frac{\partial}{\partial y} yz f'(xyz) = z f'(xyz) + yz xz f''(xyz)$$ by the product rule.

Finally, again by product and chain rule, $$\frac{\partial}{\partial z} \left(z f'(xyz) + yz xz f''(xyz)\right)= 1f'(xyz)+zxyf''(xyz)+xy2zf''(xyz)+x^2y^2z^2 f'''(xyz).$$

Putting the pieces together, we obtain $$\frac{\partial^3}{\partial x \partial y \partial z}f(xyz) =f'(xyz)+3zxyf''(xyz)+x^2y^2z^2 f'''(xyz)=F(x,y,z).$$

Hope this helps.