Problem 34 - Calculus Early Transcendentals

Question

Find the four second partial derivatives of the following functions. $$H(x, y)=\sqrt{4+x^{2}+y^{2}}$$

Step-by-Step Solution

Answer

Question: Find the four second partial derivatives of the function H(x, y) = $$\sqrt{4+x^2+y^2}$$. Answer: The four second partial derivatives of the function H(x, y) are: $$\frac{\partial^2 H}{\partial x^2} = \frac{4-y^2}{(4+x^2+y^2)^{3/2}}$$, $$\frac{\partial^2 H}{\partial x \partial y} = \frac{-xy}{(4+x^2+y^2)^{3/2}}$$, $$\frac{\partial^2 H}{\partial y \partial x} = \frac{-xy}{(4+x^2+y^2)^{3/2}}$$, $$\frac{\partial^2 H}{\partial y^2} = \frac{4-x^2}{(4+x^2+y^2)^{3/2}}$$.

Step 1: Compute the first partial derivative with respect to x

To find the first partial derivative of H(x, y) with respect to x, we can use the chain rule of differentiation as follows: $$\frac{\partial H}{\partial x} = \frac{1}{2 \sqrt{4+x^2+y^2}}\left(\frac{\partial}{\partial x} (4+x^2+y^2)\right)$$ Now, differentiate the function inside the square root with respect to x: $$\frac{\partial H}{\partial x} = \frac{1}{2 \sqrt{4+x^2+y^2}} (2x)$$ Simplify the expression: $$\frac{\partial H}{\partial x} = \frac{x}{\sqrt{4+x^2+y^2}}$$

Step 2: Compute the first partial derivative with respect to y

To find the first partial derivative of H(x, y) with respect to y, similarly use the chain rule of differentiation: $$\frac{\partial H}{\partial y} = \frac{1}{2 \sqrt{4+x^2+y^2}}\left(\frac{\partial}{\partial y} (4+x^2+y^2)\right)$$ Now, differentiate the function inside the square root with respect to y: $$\frac{\partial H}{\partial y} = \frac{1}{2 \sqrt{4+x^2+y^2}} (2y)$$ Simplify the expression: $$\frac{\partial H}{\partial y} = \frac{y}{\sqrt{4+x^2+y^2}}$$

Step 3: Compute the second partial derivatives

Now that we have the first partial derivatives, we will compute the four second partial derivatives: 1. Second derivative with respect to x squared: Differentiate $$\frac{\partial H}{\partial x}$$ with respect to x again: $$\frac{\partial^2 H}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{x}{\sqrt{4+x^2+y^2}}\right)$$ 2. Second derivative with respect to x then y: Differentiate $$\frac{\partial H}{\partial x}$$ with respect to y: $$\frac{\partial^2 H}{\partial x \partial y}=\frac{\partial}{\partial y}\left(\frac{x}{\sqrt{4+x^2+y^2}}\right)$$ 3. Second derivative with respect to y then x: Differentiate $$\frac{\partial H}{\partial y}$$ with respect to x: $$\frac{\partial^2 H}{\partial y \partial x}=\frac{\partial}{\partial x}\left(\frac{y}{\sqrt{4+x^2+y^2}}\right)$$ 4. Second derivative with respect to y squared: Differentiate $$\frac{\partial H}{\partial y}$$ with respect to y again: $$\frac{\partial^2 H}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{y}{\sqrt{4+x^2+y^2}}\right)$$

Step 4: Calculate the second partial derivatives

By computing the derivatives, we get: 1. Second derivative with respect to x squared: $$\frac{\partial^2 H}{\partial x^2} = \frac{4-y^2}{(4+x^2+y^2)^{3/2}}$$ 2. Second derivative with respect to x then y: $$\frac{\partial^2 H}{\partial x \partial y} = \frac{-xy}{(4+x^2+y^2)^{3/2}}$$ 3. Second derivative with respect to y then x: $$\frac{\partial^2 H}{\partial y \partial x} = \frac{-xy}{(4+x^2+y^2)^{3/2}}$$ 4. Second derivative with respect to y squared: $$\frac{\partial^2 H}{\partial y^2} = \frac{4-x^2}{(4+x^2+y^2)^{3/2}}$$ So, the four second partial derivatives of the function H(x, y) are: $$\frac{\partial^2 H}{\partial x^2} = \frac{4-y^2}{(4+x^2+y^2)^{3/2}}, \frac{\partial^2 H}{\partial x \partial y} = \frac{-xy}{(4+x^2+y^2)^{3/2}}, \frac{\partial^2 H}{\partial y \partial x} = \frac{-xy}{(4+x^2+y^2)^{3/2}}, \frac{\partial^2 H}{\partial y^2} = \frac{4-x^2}{(4+x^2+y^2)^{3/2}}$$