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


Given the differentiable function


Prove the equation





We get the function


And the internal function


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=



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.


=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=


We got zero as required.

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