Question

Using gradient rules Use the gradient rules of Exercise 85 to find the gradient of the following functions. $$f(x, y)=\ln \left(1+x^{2}+y^{2}\right)$$

Short answer

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

Step-by-step solution

Identify and understand the function

We are given the function $$f(x, y) = \ln \left(1 + x^2 + y^2\right).$$ The goal is to find the gradient of this function using the gradient rules.

Calculate the partial derivative with respect to x

To find the partial derivative of the function with respect to $$x$$, we differentiate it with respect to $$x$$ while keeping $$y$$ constant. $$ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \ln \left(1 + x^2 + y^2\right) $$ Now use the chain rule to differentiate: $$ \frac{\partial f}{\partial x} = \frac{1}{1+x^2+y^2} \cdot \frac{\partial}{\partial x} (1 + x^2 + y^2) = \frac{2x}{1+x^2+y^2} $$

Calculate the partial derivative with respect to y

Similarly, to find the partial derivative of the function with respect to $$y$$, we differentiate it with respect to $$y$$ while keeping $$x$$ constant. $$ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \ln \left(1 + x^2 + y^2\right) $$ Again use the chain rule to differentiate: $$ \frac{\partial f}{\partial y} = \frac{1}{1+x^2+y^2} \cdot \frac{\partial}{\partial y} (1 + x^2 + y^2) = \frac{2y}{1+x^2+y^2} $$

Form the gradient vector

Now that we have the partial derivatives with respect to $$x$$ and $$y$$, we can form the gradient vector: $$ \nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = \left(\frac{2x}{1+x^2+y^2}, \frac{2y}{1+x^2+y^2}\right) $$ The gradient of the function $$f(x, y) = \ln \left(1 + x^2 + y^2\right)$$ is $$\nabla f(x, y) = \left(\frac{2x}{1+x^2+y^2}, \frac{2y}{1+x^2+y^2}\right).$$

Key concepts

Partial Derivatives

When dealing with functions of several variables, like in our exercise where the function is \( f(x, y) = \ln(1 + x^2 + y^2) \), it's important to understand partial derivatives. A partial derivative measures how a function changes as just one of the variables varies, while others are held constant. It's like observing the effect of jogging speed on your heart rate while keeping your diet constant.

• To find the partial derivative with respect to \( x \), differentiate \( f \) treating \( y \) as a constant.
• Similarly, to find it with respect to \( y \), treat \( x \) as a constant.

For instance, the partial derivative of our function with respect to \( x \) is \( \frac{\partial f}{\partial x} = \frac{2x}{1+x^2+y^2} \). This approach gives us insight into how sensitive our function is to changes in each variable, independently. It's a crucial concept in multivariable calculus, especially when examining phenomena where variables interact.

Chain Rule

The chain rule is an essential principle in calculus, especially when dealing with composite functions. It allows us to differentiate a function that is nested within another function. In our context,

• the chain rule is used when we take the derivative of the function \( f(x, y) = \ln(1 + x^2 + y^2) \).

Here's a simple way to think about the chain rule: imagine a function inside another, like layers of an onion. To differentiate, start from the outermost layer and work your way in. When computing the partial derivative \( \frac{\partial f}{\partial x} \), we first differentiate the outer \( \ln \) function and multiply by the derivative of its inner part \((1+x^2+y^2)\). Hence, the chain rule helps us break down complex derivatives into manageable steps, making problems like these more approachable.

Gradient Vector

The gradient vector is a fascinating concept in vector calculus that indicates both the direction and rate of the fastest increase of a function. It gives you a sense of how to move through a multi-dimensional space to increase the function's value most efficiently. In our exercise, we calculated the gradient vector \( abla f(x, y) \) of the function \( f(x, y) = \ln(1 + x^2 + y^2) \). We found it to be

• \( abla f(x, y) = \left(\frac{2x}{1+x^2+y^2}, \frac{2y}{1+x^2+y^2}\right) \).

The components of this gradient are the partial derivatives with respect to each variable, \( x \) and \( y \). With this gradient vector, you can determine:

• Which direction you should move along in the \(xy\)-plane to increase your function value the fastest,
• How steep this increase is depending on the magnitude of the vector.

Understanding the gradient vector is crucial for optimization tasks, like finding maxima or minima in calculus. It's like having a roadmap that tells you where to go if you're searching for the highest peak in a mountainous landscape.