Question

Find the first partial derivatives of the function.

\({\bf{f}}({\bf{x}},{\bf{y}}) = \frac{{\bf{x}}}{{\bf{y}}}\)

Short answer

Finding the first partial derivatives of the function \(f(x,y) = \frac{x}{y}\)

Step-by-step solution

Finding limit:

We need to find the limit, if it exists, or show that the limit does not exist.

\(\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{{x^2}y{e^x}}}{{{x^4} + 4{y^2}}}\)

Let\(f(x,y) = \frac{{{x^2}y{e^x}}}{{{x^4} + 4{y^2}}}\)

First, try approaching\(f(x)\) along the line\({\bf{y}} = {\bf{0}}\)

\(f(x,0) = \frac{{{x^2}(0)({e^0})}}{{{x^4} + 4{{(0)}^2}}}\)

\( \Rightarrow {\bf{f}}({\bf{x}},{\bf{0}}) = {\bf{0}}\) (for \(x \ne 0\))

Thus, \(f(x,y) \to 0\) along the line \({\bf{y}} = {\bf{0}}\)

Calculations :

Next, try approaching\(f(x)\) along the curve\({\bf{y}} = {{\bf{x}}^{\bf{2}}}\)

Notice that this passes through the point\(({\bf{0}},{\bf{0}})\) as required.

\(\therefore f(x,{x^2}) = \frac{{{x^2}({x^2})({e^{{x^2}}})}}{{{x^4} + 4{{({x^2})}^2}}}\) use\(f(x,y) = \frac{{{x^2}y{e^x}}}{{{x^4} + 4{y^2}}}\)

\( = \frac{{{x^4}{e^{{x^2}}}}}{{({x^4} + 4{x^4})}}\)

\( = \frac{{{x^4}{e^{{x^2}}}}}{{5{x^4}}}\)

\( = \frac{{{e^{{x^2}}}}}{5}\)

\( \Rightarrow {\bf{f}}({\bf{x}},{{\bf{x}}^{\bf{2}}}) = \frac{{{{\bf{e}}^{{{\bf{x}}^{\bf{2}}}}}}}{{\bf{5}}}\) (for \(x \ne 0\))

Thus,\(f(x,y) \to \frac{{{e^{{0^2}}}}}{5} = \frac{1}{5}\)along the curve\({\bf{y}} = {{\bf{x}}^{\bf{2}}}\).

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

Hence, the limit does not exist.