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Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6520

Exercise

Given the differentiable function

z=xf(x+y)+yg(x+y)

Prove the equation

z''_{xx}-2z''_{xy}+z''_{yy}=0

Proof

Define

u=x+y

We get the function

z=xf(u)+yg(u)

And the internal function

u(x,y)=x+y

We will use the chain rule to calculate the first and second order partial derivatives of z.

z'_x=1\cdot f+x\cdot f'_x+y\cdot g'_x=

=f+x\cdot f'_u\cdot u'_x+y\cdot g'_u\cdot u'_x=

=f+x\cdot f'_u\cdot 1+y\cdot g'_u\cdot 1=

=f+xf'_u+yg'_u=

 

z''_{xx}=f'_u\cdot u'_x+f'_u+x\cdot f''_u\cdot u'_x+y\cdot g''_u\cdot u'_x=

=f'_u\cdot 1+f'_u+x\cdot f''_u\cdot 1+y\cdot g''_u\cdot 1=

=f'_u+f'_u+xf''_u+y\cdot g''_u

 

z''_{xy}=f'_u\cdot u'_y+x\cdot f''_u\cdot u'_y+g'_u+y\cdot g''_u\cdot u'_y=

=f'_u\cdot 1+x\cdot f''_u\cdot 1+g'_u+y\cdot g''_u\cdot 1=

=f'_u+x\cdot f''_u+g'_u+y\cdot g''_u

 

z'_y=x\cdot f'_u\cdot u'_y+g+y\cdot g'_u\cdot u'_y=

=x\cdot f'_u\cdot 1+g+y\cdot g'_u\cdot 1=

=x\cdot f'_u+g+y\cdot g'_u

 

z''_{yy}=x\cdot f''_u\cdot u'_y+g'_u\cdot u'_y+g'_u+y\cdot g''_u\cdot u'_y=

=x\cdot f''_u\cdot 1+g'_u\cdot 1+g'_u+y\cdot g''_u\cdot 1=

=x\cdot f''_u+g'_u+g'_u+y\cdot g''_u

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

z''_{xx}-2z''_{xy}+z''_{yy}=

=f'_u+f'_u+xf''_u+y\cdot g''_u-2(f'_u+x\cdot f''_u+g'_u+y\cdot g''_u)+x\cdot f''_u+g'_u+g'_u+y\cdot g''_u=

=f'_u+f'_u+xf''_u+y\cdot g''_u-2f'_u-2x\cdot f''_u-2g'_u-2y\cdot g''_u+x\cdot f''_u+g'_u+g'_u+y\cdot g''_u=

=0

We got zero as required.

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