Problem 84 - Calculus Early Transcendentals

Question

Use the gradient rules of Exercise 81 to find the gradient of the following functions. $$f(x, y)=\ln \left(1+x^{2}+y^{2}\right)$$

Step-by-Step Solution

Answer

Answer: The gradient of the function $$f(x, y)=\ln \left(1+x^{2}+y^{2}\right)$$ is: $$\nabla f(x, y) = \begin{bmatrix} \frac{2x}{1+x^2+y^2} \\ \frac{2y}{1+x^2+y^2} \end{bmatrix}$$

Step 1: Identify the function and variables

We are given the function $$f(x, y)=\ln \left(1+x^{2}+y^{2}\right)$$, and we want to find its gradient. The variables involved are x and y.

Step 2: Find the partial derivative with respect to x

Using the chain rule, we find the partial derivative of the function with respect to x: $$\frac{\partial f}{\partial x} = \frac{1}{1+x^2+y^2} \cdot \frac{\partial}{\partial x}(1+x^2+y^2)$$ The derivative of \((1+x^2+y^2)\) with respect to x is \(2x\). Therefore, $$\frac{\partial f}{\partial x} = \frac{2x}{1+x^2+y^2}$$

Step 3: Find the partial derivative with respect to y

Next, we find the partial derivative of the function with respect to y: $$\frac{\partial f}{\partial y} = \frac{1}{1+x^2+y^2} \cdot \frac{\partial}{\partial y}(1+x^2+y^2)$$ The derivative of \((1+x^2+y^2)\) with respect to y is \(2y\). Therefore, $$\frac{\partial f}{\partial y} = \frac{2y}{1+x^2+y^2}$$

Step 4: Compute the gradient

Now that we have the partial derivatives with respect to x and y, we can write down the gradient of the function. The gradient is a vector consisting of the partial derivatives: $$\nabla f(x, y) = \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix} = \begin{bmatrix} \frac{2x}{1+x^2+y^2} \\ \frac{2y}{1+x^2+y^2} \end{bmatrix}$$ And that's the final answer! The gradient of the function $$f(x, y)=\ln \left(1+x^{2}+y^{2}\right)$$ is: $$\nabla f(x, y) = \begin{bmatrix} \frac{2x}{1+x^2+y^2} \\ \frac{2y}{1+x^2+y^2} \end{bmatrix}$$