Question

find all the second - order partial derivatives of the function ( f(x,y)=5x^{2}+4y + 7x^{2}y^{2} ).
( \frac{partial^{2}f}{partial x^{2}}=square )

Explanation:

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

We use the power rule \(\frac{\partial}{\partial x}(x^n)=nx^{n - 1}\) and treat \(y\) as a constant.
For \(f(x,y)=5x^{2}+4y + 7x^{2}y^{2}\), \(\frac{\partial f}{\partial x}=\frac{\partial}{\partial x}(5x^{2})+\frac{\partial}{\partial x}(4y)+\frac{\partial}{\partial x}(7x^{2}y^{2})\)
\(\frac{\partial f}{\partial x}=5\times2x+0 + 7\times2xy^{2}=10x + 14xy^{2}\)

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

Differentiate \(\frac{\partial f}{\partial x}=10x + 14xy^{2}\) with respect to \(x\) again.
\(\frac{\partial^{2}f}{\partial x^{2}}=\frac{\partial}{\partial x}(10x)+\frac{\partial}{\partial x}(14xy^{2})\)
Using the power rule \(\frac{\partial}{\partial x}(ax^{n})=anx^{n - 1}\) (\(a = 10,n = 1\) and \(a = 14y^{2},n = 1\)), we get \(\frac{\partial^{2}f}{\partial x^{2}}=10+14y^{2}\)

Answer:

\(10 + 14y^{2}\)