Question

Show the function \(f(x,y) = {x^2} + {y^2}\)is differentiable by obtaining the values of \({\varepsilon _1}\)and \({\varepsilon _2}\)by using Definition 7.

Short answer

The answer is stated below.

Step-by-step solution

Definition 7.

“If\(z = f(x,y){\rm{,}}\)then\(f\)is differentiable at\((a,b)\)if\(\Delta z\)can be expressed in the form\(\Delta z = {f_x}(a,b)\Delta x + {f_y}(a,b)\Delta y + {\varepsilon _1}\Delta x + {\varepsilon _2}\Delta y,\)where\({\varepsilon _1}\)and\({\varepsilon _2} \to 0.{\rm{ as }}(\Delta x,\Delta y) \to (0,0).\)”

Take the partial derivatives for calculation.

Consider the function \(z = f(x,y) = {x^2} + {y^2}.\)

Take the partial derivatives with respect to \(x\) and \(y\)in the function\(z = f(x,y) = {x^2} + {y^2}.\)

\(\begin{aligned}{l}{f_x}\left( {{x^2} + {y^2}} \right) = 2x + 0\\ = 2x\\{f_y}\left( {{x^2} + {y^2}} \right) = 0 + 2y\\ = 2y\end{aligned}\)

Thus, the value of\({f_x}(x,y) = 2x\)and\({f_y}(x,y) = 2y\).

Compute the values of\({\varepsilon _1}\) and\({\varepsilon _2}\) by using Definition 7 as follows\(\Delta z = {f_x}(x,y)\Delta x + {f_y}(x,y)\Delta y + {\varepsilon _1}\Delta x + {\varepsilon _2}\Delta y\) …… (1)

The value of\(\Delta z\) can be expressed as follows.

\(\begin{aligned}{l}\Delta z = f(x + \Delta x,y + \Delta y) - f(x,y)\\ = \left( {{{(x + \Delta x)}^2} + {{(y + \Delta y)}^2}} \right) - \left( {{x^2} + {y^2}} \right)\\ = \left( {{x^2} + 2x\Delta x + \Delta {x^2} + {y^2} + 2y\Delta y + \Delta {y^2}} \right) - \left( {{x^2} + {y^2}} \right)\\ = 2x\Delta x + \Delta {x^2} + 2y\Delta y + \Delta {y^2}\end{aligned}\)

Simplify further as follows.

\(\begin{aligned}{l}\Delta z = 2x\Delta x + \Delta {x^2} + 2y\Delta y + \Delta {y^2}\\ = 2x\Delta x + 2y\Delta y + \Delta {x^2} + \Delta {y^2}\\ = 2x\Delta x + 2y\Delta y + \Delta x\Delta x + \Delta y\Delta y\end{aligned}\)

Thus, the value of\(\Delta z = 2x\Delta x + 2y\Delta y + \Delta x\Delta x + \Delta y\Delta y\) …… (2)

Compare the equation (1) and (2) and obtain the value of\({\varepsilon _1}\)and\({\varepsilon _2}\),

\(2x\Delta x + 2y\Delta y + \Delta x\Delta x + \Delta y\Delta y = {f_x}(x,y)\Delta x + {f_y}(x,y)\Delta y + {\varepsilon _1}\Delta x + {\varepsilon _2}\Delta \)

\(y{\varepsilon _1} = \Delta x\)and\({\varepsilon _2} = \Delta y.\)

Thus, the value of\(y{\varepsilon _1} = \Delta x\)and \({\varepsilon _2} = \Delta y.\)

Hence, the function\(f(x,y) = {x^2} + {y^2}\) is differentiable.