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Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 3384

Exercise

Given the differentiable function

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

Prove the equation

u'_x+u'_y+u'_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

v=x-y

w=y-z

t=z-x

We get the function

u(x,y,z)=f(v,w,t)

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

u'_x=f'_v\cdot v'_x+f'_w\cdot w'_x+f'_t\cdot t'_x

u'_y=f'_v\cdot v'_y+f'_w\cdot w'_y+f'_t\cdot t'_y

u'_z=f'_v\cdot v'_z+f'_w\cdot w'_z+f'_t\cdot t'_z

We calculate  the partial derivatives of v,w and t.

v'_x=1

v'_y=-1

v'_z=0

w'_x=0

w'_y=1

w'_z=-1

t'_x=-1

t'_y=0

t'_z=1

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

u'_x=f'_v\cdot v'_x+f'_w\cdot w'_x+f'_t\cdot t'_x=

=f'_v\cdot 1+f'_w\cdot 0+f'_t\cdot (-1)=

=f'_v-f'_t

u'_y=f'_v\cdot -1+f'_w\cdot w'_y+f'_t\cdot t'_y=

=f'_v\cdot (-1)+f'_w\cdot 1+f'_t\cdot 0=

=-f'_v+f'_w

u'_z=f'_v\cdot v'_z+f'_w\cdot w'_z+f'_t\cdot t'_z=

=f'_v\cdot 0+f'_w\cdot (-1)+f'_t\cdot 1=

=-f'_w+f'_t

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

u'_x+u'_y+u'_z=

=(f'_v-f'_t)+(-f'_v+f'_w)+(-f'_w+f'_t)=

=f'_v-f'_t-f'_v+f'_w-f'_w+f'_t=0

Hence, we got

u'_x+u'_y+u'_z=0

As required.

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