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.

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!