Question

find ( f_{x}(x,y) ) and ( f_{y}(x,y) ). then, find ( f_{x}(3,4) ) and ( f_{y}(4,2) ).
( f(x,y)=7xy + 6y^{4}+2 )
( f_{x}(x,y)= )

Explanation:

Step1: Find \(f_x(x,y)\)

Differentiate \(f(x,y)=7xy + 6y^{4}+2\) with respect to \(x\) (treating \(y\) as a constant).
Using the power rule \(\frac{\partial}{\partial x}(ax^{n}y^{m})=nax^{n - 1}y^{m}\) (here \(n = 1\) for the \(xy\) term), \(\frac{\partial}{\partial x}(7xy)=7y\), \(\frac{\partial}{\partial x}(6y^{4}) = 0\) (since \(y\) is treated as a constant), \(\frac{\partial}{\partial x}(2)=0\). So \(f_x(x,y)=7y\).

Step2: Find \(f_y(x,y)\)

Differentiate \(f(x,y)=7xy + 6y^{4}+2\) with respect to \(y\) (treating \(x\) as a constant).
Using the power rule \(\frac{\partial}{\partial y}(ax^{n}y^{m})=max^{n}y^{m - 1}\), \(\frac{\partial}{\partial y}(7xy)=7x\), \(\frac{\partial}{\partial y}(6y^{4})=24y^{3}\), \(\frac{\partial}{\partial y}(2)=0\). So \(f_y(x,y)=7x + 24y^{3}\).

Step3: Find \(f_x(3,4)\)

Substitute \(x = 3\) and \(y = 4\) into \(f_x(x,y)\). Since \(f_x(x,y)=7y\), then \(f_x(3,4)=7\times4=28\).

Step4: Find \(f_y(4,2)\)

Substitute \(x = 4\) and \(y = 2\) into \(f_y(x,y)\). Since \(f_y(x,y)=7x + 24y^{3}\), then \(f_y(4,2)=7\times4+24\times2^{3}=28 + 24\times8=28+192 = 220\).

Answer:

\(f_x(x,y)=7y\), \(f_y(x,y)=7x + 24y^{3}\), \(f_x(3,4)=28\), \(f_y(4,2)=220\)