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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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