Question

Find the first partial derivatives for the function \(f(x,y) = {x^y}\)

Short answer

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

Step-by-step solution

Step 1:- Consider the following limit:

\(\,\mathop {lim}\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + y6}}\)

The graph of function\(z = f(x,y) = \frac{{x{y^3}}}{{{x^2} + {y^6}}}\)in the window

\((0,0.001) \times (0,0.1) \times (0,0.5)\) is as shown below

Step 2:- Finding the \(\mathop {lim}\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}}:\)

From the graph along the curve\(y = {x^{\frac{1}{3}}}\)as\((x,y) \to (0,0)\), the function constant\(z = 0.5\)

The limit of this function is\({\bf{0}}.{\bf{5}}\)

Now, \(\begin{aligned}{c}\mathop {lim}\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}} = \mathop {lim}\limits_{x \to 0} \frac{{x{{(x\frac{1}{3})}^3}}}{{{x^2} + {{({x^{\frac{1}{3}}})}^6}}}\\\mathop { = lim}\limits_{x \to 0} \frac{{{x^2}}}{{{x^2} + {x^2}}}\end{aligned}\)

\(\begin{aligned}{l} \Rightarrow \mathop {lim}\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}} &= \mathop {lim}\limits_{x \to 0} \frac{{{x^2}}}{{2{x^2}}}\\ &= \mathop {lim}\limits_{x \to 0} \frac{1}{2}\\ &= \frac{1}{2}\\\therefore \mathop {lim}\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}} &= 0.5\end{aligned}\)

Step 3:- Plotting graph and Finding limit:

The graph of function\(z = f(x,y) = \frac{{x{y^3}}}{{{x^2} + {y^6}}}\)in the window\((0,0.00001) \times (0,0.00001) \times (0,4 \times 1{0^{ - 9}})\)is as shown below

From the graph along the line\({\bf{y}} = {\bf{x}}\)as\((x,y) \to (0,0)\)the function is constant

\(z = 0\)

The limit of this function is 0.

\(\mathop {lim}\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^6}}} = \mathop {lim}\limits_{x \to 0} \frac{{x{{(x)}^3}}}{\begin{aligned}{l}{x^2} + {(x)^6}\\ &= \mathop {lim}\limits_{x \to 0} \frac{{{x^4}}}{\begin{aligned}{l}{x^2} + {x^6}\\ &= \mathop {lim}\limits_{x \to 0} \frac{{{x^4}}}{\begin{aligned}{l}{x^2}(1 + {x^4})\\ &= \frac{{{{(0)}^2}}}{\begin{aligned}{l}1 + {(0)^4}\\ = 0\\ &= \mathop {lim}\limits_{(x,y) \to (0,0)} \frac{{x{y^3}}}{{{x^2} + {y^4}}} &= 0\end{aligned}}\end{aligned}}\end{aligned}}\end{aligned}}\)

The limit of this function is not independent of path passing through\((0,0)\)and\((x,y) \to (0,0)\)

The limit of function along the different paths\((y = x\)and\(y = {x^{\frac{1}{3}}})\)is different.

The limit of this function does not exist at the point\((x,y) = (0,0)\)

Hence, the limit does not exist.