Question

let ( z = f(x,y) = 12x^{2}-16xy + 9y^{2} ). find the following using the formal definition of the partial derivative.
a. ( \frac{partial z}{partial x} )
b. ( \frac{partial z}{partial y} )
c. ( \frac{partial f}{partial x}(2,1) )
d. ( f_{y}(2,3) )

Explanation:

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

For \(z = f(x,y)=12x^{2}-16xy + 9y^{2}\), using the power rule \(\frac{\partial}{\partial x}(ax^{n}y^{m})=nax^{n - 1}y^{m}\) (where \(m\) is a constant).
\(\frac{\partial z}{\partial x}=\frac{\partial}{\partial x}(12x^{2})-\frac{\partial}{\partial x}(16xy)+\frac{\partial}{\partial x}(9y^{2})\)
\(\frac{\partial z}{\partial x}=24x-16y+0 = 24x-16y\)

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

Using the power rule \(\frac{\partial}{\partial y}(ax^{n}y^{m})=max^{n}y^{m - 1}\) (where \(n\) is a constant).
\(\frac{\partial z}{\partial y}=\frac{\partial}{\partial y}(12x^{2})-\frac{\partial}{\partial y}(16xy)+\frac{\partial}{\partial y}(9y^{2})\)
\(\frac{\partial z}{\partial y}=0-16x + 18y=18y-16x\)

Step3: Evaluate \(\frac{\partial f}{\partial x}\) at \((2,1)\)

Substitute \(x = 2\) and \(y = 1\) into \(\frac{\partial z}{\partial x}=24x-16y\)
\(\frac{\partial f}{\partial x}(2,1)=24\times2-16\times1=48 - 16=32\)

Step4: Evaluate \(f_y\) at \((2,3)\)

Since \(f_y=\frac{\partial z}{\partial y}=18y-16x\), substitute \(x = 2\) and \(y = 3\)
\(f_y(2,3)=18\times3-16\times2=54 - 32 = 22\)

Answer:

a. \(24x-16y\)
b. \(18y - 16x\)
c. \(32\)
d. \(22\)