Exercise
Given the differentiable function
z(x,y)=x^y
Prove the equation
z''_{xy}=z''_{yx}
Proof
We will use the chain rule to calculate the partial derivatives of z.
z'_x=yx^{y-1}
z'_y=x^y\ln x
We will calculate the second order derivatives.
z''_{xy}=x^{y-1}+y\cdot\frac{1}{x}\cdot x^y\cdot\ln x=
=x^{y-1}(1+y\cdot\ln x)
z''_{yx}=yx^{y-1}\cdot\ln x+x^y\cdot\frac{1}{x}
=x^{y-1}(y\cdot\ln x+1)
Hence, we got
z''_{xy}=z''_{yx}
As required.
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