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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