Problem 31 - Calculus with Concepts in Calculus

Question

Find \(f_{x y}\) and \(f_{y x}\) $$ f(x, y)=\sqrt{x^{2}+y^{2}} $$

Step-by-Step Solution

Answer

Both \( f_{xy} \) and \( f_{yx} \) are \( \frac{-xy}{(x^2+y^2)^{3/2}} \).

Step 1: Compute the partial derivative of f with respect to x

The function is given as \( f(x, y) = \sqrt{x^{2}+y^{2}} \). First, let's find the partial derivative of \( f \) with respect to \( x \). We'll use the chain rule here. Set \( u = x^2 + y^2 \) so that \( f(x, y) = \sqrt{u} = u^{1/2} \).The derivative of \( u^{1/2} \) with respect to \( u \) is \( \frac{1}{2}u^{-1/2} \).Then, multiply by the derivative of \( u \) with respect to \( x \), which is \( 2x \).Thus, \( f_x = \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + y^2}} \).

Step 2: Compute the partial derivative of f_x with respect to y

Now, take the derivative of \( f_x = \frac{x}{\sqrt{x^2 + y^2}} \) with respect to \( y \).We'll apply the quotient rule: Given a function \( g(y) = \frac{x}{h(y)} \) where \( h(y) = \sqrt{x^2 + y^2} \), then its derivative is:\[ g'(y) = \frac{0 \cdot h(y) - x \cdot \frac{d(h(y))}{dy}}{(h(y))^2} \]The derivative \( \frac{d(h(y))}{dy} = \frac{y}{\sqrt{x^2 + y^2}} \). Thus:\[ f_{xy} = \frac{-x \cdot \frac{y}{\sqrt{x^2 + y^2}}}{(\sqrt{x^2 + y^2})^2} = \frac{-xy}{(x^2+y^2)^{3/2}} \].

Step 3: Compute the partial derivative of f with respect to y

Let's calculate the partial derivative of \( f(x, y) = \sqrt{x^2 + y^2} \) with respect to \( y \).As before, set \( u = x^2 + y^2 \), then \( f(x, y) = \sqrt{u} = u^{1/2} \).Using the chain rule:The derivative of \( u^{1/2} \) with respect to \( u \) is \( \frac{1}{2}u^{-1/2} \).Multiply by the derivative of \( u \) with respect to \( y \), which is \( 2y \).Hence, \( f_y = \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot 2y = \frac{y}{\sqrt{x^2 + y^2}} \).

Step 4: Compute the partial derivative of f_y with respect to x

Now, differentiate \( f_y = \frac{y}{\sqrt{x^2+y^2}} \) with respect to \( x \).Again, using the quotient rule:Set \( j(x) = \frac{y}{h(x)} \) where \( h(x) = \sqrt{x^2 + y^2} \).\( j'(x) = \frac{0 \cdot h(x) - y \cdot \frac{d(h(x))}{dx}}{(h(x))^2} \)Applying the derivative \( \frac{d(h(x))}{dx} = \frac{x}{\sqrt{x^2 + y^2}} \):\[ f_{yx} = \frac{-y \cdot \frac{x}{\sqrt{x^2 + y^2}}}{(\sqrt{x^2 + y^2})^2} = \frac{-xy}{(x^2+y^2)^{3/2}} \].

Key Concepts

Chain Rule

The chain rule is a fundamental tool used in calculus, particularly when dealing with functions of multiple variables like in this example. It allows us to find the derivative of a composite function. If you have a function expressed in terms of another function, the chain rule helps "chain" these derivatives together.

For instance, in our given function \( f(x, y) = \sqrt{x^2 + y^2} \), we first set \( u = x^2 + y^2 \), making our function \( f(x, y) = \sqrt{u} = u^{1/2} \). To differentiate \( f \) with respect to \( x \) or \( y \), we look at two separate derivatives:

• The derivative of \( u^{1/2} \) with respect to \( u \), which gives us \( \frac{1}{2}u^{-1/2} \).
• The derivative of \( u \) with respect to \( x \) or \( y \), which is \( 2x \) or \( 2y \). Thus, to differentiate \( f \) with respect to \( x \), we multiply these derivatives: \( \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + y^2}} \).

This systematic approach allows us to manage complex derivatives efficiently.

Quotient Rule

The quotient rule is another essential technique used in calculus that helps differentiate functions expressed as one function divided by another. It’s particularly useful when the function takes the form of a fraction.

For differentiating \( f_x = \frac{x}{\sqrt{x^2 + y^2}} \) with respect to \( y \), we apply the quotient rule. The formula for the quotient rule says that for a function \( \frac{u}{v} \), the derivative is given by:

• \[ \frac{d}{dy} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dy} - u \cdot \frac{dv}{dy}}{v^2} \]
• In our case, \( u = x \) and \( v = \sqrt{x^2 + y^2} \). Thus, the derivative \( f_{xy} \) becomes \( \frac{-x \cdot \frac{y}{\sqrt{x^2+y^2}}}{(x^2+y^2)^{3/2}} \).

This rule streamlines the differentiation of fractions by utilizing both the numerator and denominator derivatives, simplifying complex operations.

Multivariable Calculus

Multivariable calculus expands on single-variable calculus by dealing with functions that depend on more than one variable. This branch of calculus introduces concepts like partial derivatives, which measure how a function changes as only one variable at a time is varied.

In the context of our exercise with \( f(x, y) = \sqrt{x^2 + y^2} \), we explore partial derivatives. A partial derivative of a function of two variables \( f(x, y) \) with respect to \( x \) is represented as \( f_x \), and with respect to \( y \) as \( f_y \).

• The process involves taking derivatives with respect to one variable while keeping others constant.
• For \( f_{xy} \) and \( f_{yx} \), we first find \( f_x \) and \( f_y \), then differentiate these with respect to the other variable.

Using multivariable calculus, we effectively explore how small changes in one direction impact the function's behavior across different dimensions, giving a deeper understanding of their dynamics.