Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 3381

Exercise

Given the differentiable function

$u(x,y,z)=f(x^2z-yz)$

Prove the equation

$xu'_x+2yu'_y-2zu'_z=0$

Proof

When we have a function and in its parentheses there is a complex expression instead of a simple variable, we will define a new variable like this

$t=x^2z-yz$

We get the function

$u(x,y,z)=f(t)$

Or simply

$u(x,y,z)=f$

Now we will use the chain rule to calculate the partial derivatives of z.

$u'_x=f'_t\cdot t'_x$

$u'_y=f'_t\cdot t'_y$

$u'_z=f'_t\cdot t'_z$

We will calculate the partial derivatives of t.

$t'_x=2xz$

$t'_y=-z$

$t'_z=x^2-y$

We put the results in the partial derivatives of u and get

$u'_x=f'_t\cdot t'_x=f'_t\cdot 2xz$

$u'_y=f'_t\cdot t'_y=f'_t\cdot (-z)$

$u'_z=f'_t\cdot t'_z=f'_t\cdot (x^2-y)$

We will put the partial derivatives in the left side of the equation we need to prove.

$xu'_x+2yu'_y-2zu'_z=$

$=xf'_t\cdot 2xz+2yf'_t\cdot (-z)-2zf'_t\cdot (x^2-y)=$

$=2zx^2f'_t-2yzf'_t-2zx^2f'_t +2yzf'_t=0$

Hence, we get

$xu'_x+2yu'_y-2zu'_z=0$

As required.

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