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Chapter 13: Problem 84

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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, $$\frac{\partial f}{\partial x} = \frac{2x}{1+x^2+y^2}$$
03

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, $$\frac{\partial f}{\partial y} = \frac{2y}{1+x^2+y^2}$$
04

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: $$\nabla f(x, y) = \begin{bmatrix} \frac{2x}{1+x^2+y^2} \\ \frac{2y}{1+x^2+y^2} \end{bmatrix}$$

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