Question

Find the partial derivatives of the function.

\(g(u,v) = {({u^2}v - {v^3})^5}\)

Short answer

Finding the first partial derivatives of the function \(g(u,v) = {({u^2}v - {v^3})^5}\)

Step-by-step solution

Consider the following limit: 

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

The objective is to verify that the limit exists or not.

Let the function be c

First let’s approach \((0,0,0)\) along the non-vertical line through the origin.

Then \(y = z = 0\)

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

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

, (for all \(x \ne 0\))

Thus, \(f(x,y,z) \to 0\)along the \(x\)-axis.

Calculation:

Let approach \((0,0,0)\)along the \(y\)-axis.

Then \(x = z = 0\)

\(f(0,y,0) = \frac{{y(0) + y0 + x(0)}}{{0 + {y^2} + 0}}\)Use \(f(0,y,0) = \frac{{y(0) + y0 + x(0)}}{{0 + {y^2} + 0}}\)

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

, (for all \(y \ne 0\))

Thus, \(f(x,y,z) \to 0\)along the \(y\)-axis.

Lets approach \((0,0,0)\)along the z-axis.

Then \(x = y = 0\)

\(f(0,0,z) = \frac{{0 + z0 + z(0)}}{{0 + 0 + {z^2}}}\)

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

for all \(z \ne 0\)

Thus, \(f(x,y,z) \to 0\)along the \(z\)-axis.

let’s approach \((0,0,0)\) along another line \(y = mx\), \(z = 0\) for \(x \ne 0\)

Then, \(f(x,mx,0) = \frac{{m{x^2}}}{{{x^2}(1 + {m^2})}}\)

Which will have different values for different \(m\), for \(x \ne 0\).

Thus, the function \(f(x,y,z)\) approaches different limits along different paths.

Therefore, the limit \(\mathop {\lim }\limits_{(x,y,z) \to (0,0,0)} \frac{{xy + y{z^2} + x{z^2}}}{{{x^2} + {y^2} + {z^4}}}\)does not exist.