Problem 13 - The Calculus with Analytic Geometry

Question

In Exercises 13 through 24 , find the indicated partial derivatives by holding all but one of the variables constant and applying theorems for ordinary differentiation. $$ f(x, y)=4 y^{3}+\sqrt{x^{2}+y^{2}} ; D_{1} f(x, y) $$

Step-by-Step Solution

Answer

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

Step 1 - Identify the partial derivative

We need to find the partial derivative of the function \(f(x, y) = 4y^{3} + \sqrt{x^{2} + y^{2}}\) with respect to \(x\), denoted as \(D_{1}f(x, y)\) or \(\frac{\partial f}{\partial x}\).

Step 2 - Recognize terms dependent on x

Identify the terms in the function that are dependent on \(x\). Here, the term containing \(x\) is \(\sqrt{x^{2} + y^{2}}\). The term \(4y^{3}\) does not involve \(x\) and can be treated as a constant.

Step 3 - Apply the chain rule

To differentiate \(\sqrt{x^{2} + y^{2}}\) with respect to \(x\), use the chain rule. Let \(u = x^{2} + y^{2}\), so \(\sqrt{u}\) is differentiated as \(\frac{d}{dx} (u^{1/2})\).

Step 4 - Differentiate the inner function u

Differentiate \(u = x^{2} + y^{2}\) with respect to \(x\). This yields \(\frac{du}{dx} = 2x\).

Step 5 - Combine results using chain rule

According to the chain rule, the differentiation of \(\sqrt{u}\) where \(u = x^{2} + y^{2}\) is \(\frac{d}{dx} (u^{1/2}) = \frac{1}{2} u^{-1/2} \frac{du}{dx} = \frac{1}{2}(x^{2} + y^{2})^{-1/2}(2x)\).

Step 6 - Simplify the expression

Simplifying the expression \(\frac{1}{2}(x^{2} + y^{2})^{-1/2}(2x)\), we get \(\frac{x}{\sqrt{x^{2} + y^{2}}}\).

Step 7 - Combine final result

Considering only non-zero terms dependent on \(x\), we find \(D_{1}f(x, y) = \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^{2} + y^{2}}}\).

Key Concepts

chain rule

The chain rule is a fundamental concept in calculus used for computing the derivative of a composite function. If you have two functions, say, \( g(u) \) and \( u=f(x) \), the chain rule helps in differentiating the composite function \( g(f(x)) \). To apply the chain rule, you follow these steps:

• Differentiating the outer function \( g(u) \) with respect to the intermediate variable \( u \).
• Then, differentiating the inner function \( f(x) \) with respect to the original variable \( x \).
• Multiplying these derivatives together gives the derivative of the composite function: \( \frac{d}{dx}[g(f(x))]= g' (f(x)) * f' (x) \) .

In our exercise, we used the chain rule to differentiate the term \( \sqrt{x^{2}+y^{2}} \) with respect to \( x \). Here, the inner function is \( u = x^{2} + y^{2} \) , and the outer function is \( \sqrt{u} \) . Following the application of the chain rule, we first differentiate the outer function and later the inner function.

ordinary differentiation

Ordinary differentiation involves finding the rate at which a function changes with respect to one of its variables. It primarily deals with functions of one variable. For instance, if you have a function \( f(x) \), its ordinary derivative \( \frac{df}{dx} \) indicates how \( f \) changes as \( x \) changes. In our exercise, ordinary differentiation is used when finding the partial derivative of a multi-variable function, but treating other variables as constants. For example, to find \( \frac{\partial}{\partial x}f(x, y) \), we differentiate with respect to \( x \) and treat \( y \) as a constant, effectively transforming the function temporarily into a single-variable function.

function of multiple variables

A function of multiple variables takes more than one input. For example, \( f(x, y) \) takes two inputs, \( x \) and \( y \). The output value of \( f(x, y) \) depends on both these inputs. These types of functions e increasingly common in fields such as physics and economics. Key aspects to consider with multivariable functions include:

• Each variable can independently influence the function's output.
• You can fix one variable and investigate the function's behavior with respect to the other variable.
• Common visualization tools include 3D plots, where the third dimension represents the function's output.

In our exercise, the function \( f(x, y) = 4y^{3} + \sqrt{x^{2} + y^{2}} \) is a function of two variables. To find partial derivatives, we observe how the function changes with respect to one variable, keeping the other constant.

partial differentiation steps

Partial differentiation involves finding the derivative of a function of multiple variables with respect to one of these variables. The steps to do this include:

• Identify the variable with respect to which you are differentiating.
• Treat all other variables as constants.
• Apply the differentiation rules (product rule, chain rule, etc.) as if it were a single-variable function.
• Simplify the resulting expression.

Let's revisiting our exercise:

• We started by identifying the partial derivative \( \frac{\partial f}{\partial x} \) .
• Recognized the terms that depend on \( x \) -- namely, \( \sqrt{x^{2} + y^{2}} \).
• Other terms like \( 4y^{3} \) were treated as constants.
• Next, with the chain rule, differentiated \( \sqrt{x^{2} + y^{2}} \) with respect to \( x \).
• Finally, combined and simplified the results, yielding \( \frac{x}{\sqrt{x^{2} + y^{2}}} \).

By following these steps, you accurately find how a function's output changes with one variable while holding others constant.