Question

Find the first partial derivatives of the function.

\({\bf{f}}({\bf{x}},{\bf{y}}) = {{\bf{x}}^{\bf{4}}}{{\bf{y}}^{\bf{3}}} + {\bf{8}}{{\bf{x}}^{\bf{2}}}{\bf{y}}\)

Short answer

Finding the first partial derivatives of the function \(f(x,y) = {x^4}{y^3} + 8{x^2}y\)

Step-by-step solution

The objective is to find the limit for the following expression if exist:

\(\mathop {\lim }\limits_{(x,y) \to (1,0)} \frac{{xy - y}}{{{{(x - 1)}^2} + {y^2}}}\)

Now, verify that the limit exists or not.

Let\(f(x,y) = \frac{{xy - y}}{{{{(x - 1)}^2} + {y^2}}}\)

If we substitute the limit directly then,

\(\begin{aligned}{l}\mathop {\lim }\limits_{(x,y) \to (1,0)} \frac{{xy - y}}{{{{(x - 1)}^2} + {y^2}}} &= \frac{{(1)(0) - 0}}{{{{(1 - 1)}^2} + {{(0)}^2}}}\\ &= \frac{0}{0}\end{aligned}\)

This is anindeterminateform. So evaluate this limit in the following way.

Finding the limit:

To Evaluate the limit along X-axis, substitute\(y = 0\)

Then,

\(f(x,0) = \frac{{x(0) - 0}}{{{{(x - 1)}^2} + {{(0)}^2}}}\)

\( = \frac{0}{{{{(x - 1)}^2} + {{(0)}^2}}}\)

\(\therefore f(x,y) = 0\)

Thus, \(f(x,y) \to 0\) along the line \(y = 0\) (axis).

Finding limit \(\mathop {\lim }\limits_{(x,y) \to (1,y)} \frac{{xy - y}}{{{{(x - 1)}^2} + {y^2}}}\):

To Evaluate the limit along the line parallel to y-axis, substitute\(x = 1\)

Then,

\(f(1,y) = \frac{{(1)y - y}}{{{{(1 - 1)}^2} + {{(y)}^2}}}\)

\( = \frac{{y - y}}{{0 + {{(y)}^2}}}\)

\( = \frac{0}{{{y^2}}}\)

\(\therefore f(1,y) = 0\)

Thus, along the line \(x = 1\) parallel to y-axis \(f(x,y) \to 0\)

Finding the value:

From the above resultants, the limits along the axes are identical and that do not show that the given limit is\(0\). Next, try approach\(f(x)\)along the line\(y = x - 1\)

Notice that this line passes through the point\((1,0)\)as required.

\(f(x,x - 1) = \frac{{x(x - 1) - (x - 1)}}{{{{(x - 1)}^2} + {{(x - 1)}^2}}}\)

\( = \frac{{(x - 1)(x - 1)}}{{2{{(x - 1)}^2}}}\)

\( = \frac{{{{(x - 1)}^2}}}{{2{{(x - 1)}^2}}}\)

\( = \frac{1}{2}\)

Thus,\({\bf{f}}({\bf{x}},{\bf{y}}) \to \frac{{\bf{1}}}{{\bf{2}}}\)along the line\(y = x - 1\).

As \(f\)has two different limits along two different paths, it follows that the limit does not exist.

Hence, the limit does not exist.