Question #249198

Find the partial derivative zy\displaystyle{\frac{\partial z}{\partial y}} of the function z2=x2+y\displaystyle{z^{2}=x^2+y}


1
Expert's answer
2021-10-12T06:21:06-0400

(z2)y=(x2+y)y(z^{2})'{\scriptscriptstyle y} = (x^{2}+y)'{\scriptscriptstyle y}

2zzy=12zz'{\scriptscriptstyle y} = 1

zy=zy=12z{\frac {∂z} {∂y}}=z'{\scriptscriptstyle y}= {\frac 1 {2z}}


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