# 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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