Question

Find the partial derivatives of the function.

\(w = \frac{{{e^v}}}{{u + {v^2}}}\)

Short answer

Finding the first partial derivatives of the function.

\(w = \frac{{{e^v}}}{{u + {v^2}}}\)

Step-by-step solution

Given data:

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

We have to evaluate \(\mathop {lim}\limits_{(x,y) \to (0,0)} \frac{{{x^4} - {y^4}}}{{{x^2} + {y^2}}}\)

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

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

Hence, the \(\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{{x^4} - {y^4}}}{{{x^2} + {y^2}}} = 0\).