Question

Show the function \(f(x,y) = xy - 5{y^2}\)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 \(f(x,y) = xy - 5{y^2}.\)

Take the partial derivatives with respect to \(x\) and \(y\)in the function \(f(x,y) = xy - 5{y^2}.\)

\(\begin{aligned}{l}{f_x}\left( {xy - 5{y^2}} \right) = y - 0\\ = y\\{f_y}\left( {xy - 5{y^2}} \right) = x - 5(2y)\\ = x - 10y\end{aligned}\)

Thus, the value of \({f_x}(x,y) = y\)and\({\rm{ }}{f_y}(x,y) = x - 10y\).

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.

\(\begin{aligned}{l}\Delta z = f(x + \Delta x,y + \Delta y) - f(x,y)\\ = \left( {(x + \Delta x) \cdot (y + \Delta y) - 5{{(y + \Delta y)}^2}} \right) - \left( {xy - 5{y^2}} \right)\\ = \left( {xy + x\Delta y + y\Delta x + \Delta x\Delta y - 5\left( {{y^2} + 2y\Delta y + \Delta {y^2}} \right)} \right) - \left( {xy - 5{y^2}} \right)\\ = xy + x\Delta y + y\Delta x + \Delta x\Delta y - 5{y^2} - 10y\Delta y - 5\Delta {y^2} - xy + 5{y^2}\end{aligned}\)

Simplify further as follows.

\(\begin{aligned}{l}\Delta z = xy + x\Delta y + y\Delta x + \Delta x\Delta y - 5{y^2} - 10y\Delta y - 5\Delta {y^2} - xy + 5{y^2}\\ = x\Delta y + y\Delta x + \Delta x\Delta y - 10y\Delta y - 5\Delta {y^2}\\ = y\Delta x + (x - 10y)\Delta y + \Delta x\Delta y - 5\Delta y\Delta y\end{aligned}\)

Thus, the value of \(\Delta z = y\Delta x + (x - 10y)\Delta y + \Delta x\Delta y - 5\Delta y\Delta y\) …… (2)

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

\(y\Delta x + (x - 10y)\Delta y + \Delta y\Delta x - 5\Delta y\Delta y = {f_x}(x,y)\Delta x + {f_y}(x,y)\Delta y + {\varepsilon _1}\Delta x + {\varepsilon _2}\Delta y\)

\({\varepsilon _1} = \Delta y\)and\({\varepsilon _2} = - 5\Delta y\).

Thus, the value of \({\varepsilon _1} = \Delta y\)and\({\varepsilon _2} = - 5\Delta y\).

Hence, the function\(f(x,y) = xy - 5{y^2}.\) is differentiable.