Question

Use the limit definition of partial derivatives to find \(f_{x}(x, y)\) and \(f_{y}(x, y)\). \(f(x, y)=x^{2}-2 x y+y^{2}\)

Short answer

The x partial derivative \(f_{x}(x, y)\) for the function \(f(x, y)=x^{2}-2 x y+y^{2}\) is 2x and the y partial derivative \(f_{y}(x, y)\) is -2x+2y.

Step-by-step solution

Find the x partial derivative \(f_{x}(x, y)\)

The x partial derivative \(f_{x}(x, y)\) is calculated as the derivative of \(f(x, y)\) with respect to x, treating y as a constant. Using the limit definition, this is computed as \[ f_{x}(x, y)=\lim _{h \rightarrow 0} \frac{f(x+h, y)-f(x, y)}{h} \] By substituting \(f(x, y)=x^{2}-2 x y+y^{2}\) into the equation, we get \[ f_{x}(x, y)=-\lim _{h \rightarrow 0} \frac{(x+h)^{2}-2 (x+h) y+y^{2}-x^{2}+2 x y-y^{2}}{h} \] Simplifying this gives \[ f_{x}(x, y)=\lim _{h \rightarrow 0} \frac{2 x h+h^{2}}{h}=\lim _{h \rightarrow 0}(2 x+h)=2 x \] Therefore, the x partial derivative \(f_{x}(x, y)\) for the given function is 2x.

Find the y partial derivative \(f_{y}(x, y)\)

Similarly, the y partial derivative \(f_{y}(x, y)\) is calculated as the derivative of \(f(x, y)\) with respect to y, treating x as a constant. Again, using the limit definition, this is computed as \[ f_{y}(x, y)=\lim _{h \rightarrow 0} \frac{f(x, y+h)-f(x, y)}{h} \] Substituting \(f(x, y)=x^{2}-2 x y+y^{2}\) into the equation, we get \[ f_{y}(x, y)=-\lim _{h \rightarrow 0} \frac{x^{2}-2 x(y+h)+(y+h)^{2}-x^{2}+2 x y-y^{2}}{h} \] Simplifying this gives \[ f_{y}(x, y)=\lim _{h \rightarrow 0} \frac{-2 x h+2 y h+h^{2}}{h}=\lim _{h \rightarrow 0}(-2 x+2 y+h) = -2x +2y \] Therefore, the y partial derivative \(f_{y}(x, y)\) for the given function is -2x+2y.