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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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