Problem 31 - Calculus Early Transcendentals

Question

Find the four second partial derivatives of the following functions. $$p(u, v)=\ln \left(u^{2}+v^{2}+4\right)$$

Step-by-Step Solution

Answer

Question: Calculate the four second partial derivatives of the function \(p(u, v) = \ln(u^2 + v^2 + 4)\). Answer: The four second partial derivatives are: 1. \(\frac{\partial^2 p(u, v)}{\partial u^2} = \frac{2\left(v^{2}-u^{2}+4\right)}{(u^{2}+v^{2}+4)^{2}}\) 2. \(\frac{\partial^2 p(u, v)}{\partial u \partial v} = -\frac{4uv}{(u^{2}+v^{2}+4)^{2}}\) 3. \(\frac{\partial^2 p(u, v)}{\partial v \partial u} = -\frac{4uv}{(u^{2}+v^{2}+4)^{2}}\) 4. \(\frac{\partial^2 p(u, v)}{\partial v^2} = \frac{2\left(u^{2}-v^{2}+4\right)}{(u^{2}+v^{2}+4)^{2}}\)

Step 1: Find the first partial derivatives

Compute the first partial derivatives of \(p(u, v)=\ln \left(u^{2}+v^{2}+4\right)\) with respect to \(u\) and \(v\): $$\frac{\partial p(u, v)}{\partial u} = \frac{\partial}{\partial u}\ln \left(u^{2}+v^{2}+4\right) = \frac{2u}{u^{2}+v^{2}+4}$$ $$\frac{\partial p(u, v)}{\partial v} = \frac{\partial}{\partial v}\ln \left(u^{2}+v^{2}+4\right) = \frac{2v}{u^{2}+v^{2}+4}$$

Step 2: Find the second partial derivative with respect to \(u\) twice

Differentiate the first partial derivative with respect to \(u\) again to find the second partial derivative: $$\frac{\partial^2 p(u, v)}{\partial u^2} = \frac{\partial}{\partial u} \frac{2u}{u^{2}+v^{2}+4} = \frac{2\left(u^{2}+v^{2}+4\right) - 2u(2u)}{(u^{2}+v^{2}+4)^{2}} = \frac{2\left(v^{2}-u^{2}+4\right)}{(u^{2}+v^{2}+4)^{2}}$$

Step 3: Find the second mixed partial derivative with respect to \(u\) and \(v\)

Differentiate the first partial derivative with respect to \(u\) once and then with respect to \(v\): $$\frac{\partial^2 p(u, v)}{\partial u \partial v} = \frac{\partial}{\partial v} \frac{2u}{u^{2}+v^{2}+4} = \frac{-2u(2v)}{(u^{2}+v^{2}+4)^{2}} = -\frac{4uv}{(u^{2}+v^{2}+4)^{2}}$$

Step 4: Find the second mixed partial derivative with respect to \(v\) and \(u\)

Differentiate the first partial derivative with respect to \(v\) once and then with respect to \(u\): $$\frac{\partial^2 p(u, v)}{\partial v \partial u} = \frac{\partial}{\partial u} \frac{2v}{u^{2}+v^{2}+4} = \frac{-2v(2u)}{(u^{2}+v^{2}+4)^{2}} = -\frac{4uv}{(u^{2}+v^{2}+4)^{2}}$$

Step 5: Find the second partial derivative with respect to \(v\) twice

Differentiate the first partial derivative with respect to \(v\) again to find the second partial derivative: $$\frac{\partial^2 p(u, v)}{\partial v^2} = \frac{\partial}{\partial v} \frac{2v}{u^{2}+v^{2}+4} = \frac{2\left(u^{2}+v^{2}+4\right) - 2v(2v)}{(u^{2}+v^{2}+4)^{2}} = \frac{2\left(u^{2}-v^{2}+4\right)}{(u^{2}+v^{2}+4)^{2}}$$ The four second partial derivatives are: 1. \(\frac{\partial^2 p(u, v)}{\partial u^2} = \frac{2\left(v^{2}-u^{2}+4\right)}{(u^{2}+v^{2}+4)^{2}}\) 2. \(\frac{\partial^2 p(u, v)}{\partial u \partial v} = -\frac{4uv}{(u^{2}+v^{2}+4)^{2}}\) 3. \(\frac{\partial^2 p(u, v)}{\partial v \partial u} = -\frac{4uv}{(u^{2}+v^{2}+4)^{2}}\) 4. \(\frac{\partial^2 p(u, v)}{\partial v^2} = \frac{2\left(u^{2}-v^{2}+4\right)}{(u^{2}+v^{2}+4)^{2}}\)