Find the partial derivative ∂z∂y\displaystyle{\frac{\partial z}{\partial y}}∂y∂z of the function z2=x2+y\displaystyle{z^{2}=x^2+y}z2=x2+y
(z2)′y=(x2+y)′y(z^{2})'{\scriptscriptstyle y} = (x^{2}+y)'{\scriptscriptstyle y}(z2)′y=(x2+y)′y
2zz′y=12zz'{\scriptscriptstyle y} = 12zz′y=1
∂z∂y=z′y=12z{\frac {∂z} {∂y}}=z'{\scriptscriptstyle y}= {\frac 1 {2z}}∂y∂z=z′y=2z1
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