Problem 6 - Calculus with Concepts in Calculus

Question

Find the first partial derivatives of the function. $$ f(x, y)=\sqrt{x^{2}+y^{2}} $$

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}, \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \).

Step 1: Understanding the Function

We are tasked with finding the first partial derivatives of the function \( f(x, y) = \sqrt{x^2 + y^2} \). This involves computing the derivatives with respect to each variable separately, while treating the other variable as a constant.

Step 2: Compute Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) as constant. The function can be rewritten using the power rule: \( f(x, y) = (x^2 + y^2)^{1/2} \). Using the chain rule, the derivative is: \[ \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + y^2}} \] Thus, \( \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \).

Step 3: Compute Partial Derivative with Respect to y

Now, find \( \frac{\partial f}{\partial y} \), treating \( x \) as a constant. Similarly, using the chain rule the derivative is: \[ \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot 2y = \frac{y}{\sqrt{x^2 + y^2}} \]This gives us \( \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \).

Key Concepts

Chain Rule

The chain rule is a vital concept in calculus, especially when dealing with compositions of functions. It allows us to find derivatives of composite functions, essentially functions within functions, by taking a systematic approach.

Here's how it generally works: If you have a function composed of another function, say, \( g(h(x)) \), the chain rule states that the derivative \( g'(h(x)) \)\(( h'(x) )\).

In our exercise, we're dealing with the function \( f(x, y) = \sqrt{x^{2} + y^{2}} \). We see that inside this function, we have another base expression, \( x^{2} + y^{2} \), which is raised to the power of \( 1/2 \).

When applying the chain rule while keeping one variable constant, this rule helps us break down the problem into more manageable parts, effectively distributing the derivative inside our function. This allows you to focus on one aspect at a time while ensuring that all parts are accounted for.

Power Rule

The power rule is often the first rule we learn about when differentiating functions and is simple yet powerful. It states that the derivative of \( x^{n} \) is \( nx^{n-1} \).

In the exercise, we transform the original expression under the square root into a power function: \( (x^{2} + y^{2})^{1/2} \).

Applying the power rule here results in lowering the power by one, as follows:

• Multiply by the current power, \( 1/2 \)
• Reduce the exponent by one, \( (x^{2} + y^{2})^{-1/2} \)

Combining this outcome with the chain rule, as you saw before, allows us to handle more complicated expressions efficiently. It shows the broader application of the power rule when incorporated in multivariable calculus as well.

Multivariable Calculus

Multivariable Calculus is a branch of calculus focusing on functions with more than one variable. Partial derivatives are a crucial topic here, representing how the function changes as one variable varies while the others remain fixed.

In our exercise, this is depicted through the function \( f(x, y) = \sqrt{x^{2} + y^{2}} \), where both \( x \) and \( y \) influence the outcome.

We calculate the partial derivatives \(( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} )\) by:

• Focusing on one variable while the others are treated as constants
• Applying rules like the chain and power rules to achieve the derivatives

These calculations help us understand how the function behaves in relation to each individual variable and is foundational for more advanced topics in multivariable calculus, such as gradient vectors and optimization in multi-dimensional spaces.