Question

find all the second - order partial derivatives of the function ( f(x,y)=5x^{2}+)
( \frac{partial^{2}f}{partial x^{2}}=10 + 14y^{2})
( \frac{partial^{2}f}{partial ypartial x}=28xy)
( \frac{partial^{2}f}{partial y^{2}}=)

Explanation:

Step1: Find the first - order partial derivative with respect to \(y\)

Given \(f(x,y)=5x^{2}+7x y^{2}\) (assuming the complete function is \(f(x,y)=5x^{2}+7x y^{2}\) as the original is cut off, but based on \(\frac{\partial^{2}f}{\partial x^{2}} = 10 + 14y^{2}\) and \(\frac{\partial^{2}f}{\partial y\partial x}=28xy\)).
Using the power rule for partial derivatives \(\frac{\partial f}{\partial y}=\frac{\partial(5x^{2}+7x y^{2})}{\partial y}\).
Since \(\frac{\partial(5x^{2})}{\partial y}=0\) (treating \(x\) as a constant) and \(\frac{\partial(7x y^{2})}{\partial y}=7x\times2y = 14xy\). So \(\frac{\partial f}{\partial y}=14xy\).

Step2: Find the second - order partial derivative with respect to \(y\)

Differentiate \(\frac{\partial f}{\partial y}=14xy\) with respect to \(y\) again.
Using the power rule \(\frac{\partial^{2}f}{\partial y^{2}}=\frac{\partial(14xy)}{\partial y}\).
Treating \(x\) as a constant, \(\frac{\partial(14xy)}{\partial y}=14x\).

Answer:

\(14x\)