Problem 17 - Calculus of a Single Variable

Question

Finding a Differential In Exercises \(11-20\) , find the differential \(d y\) of the given function. $$ y=\sqrt{9-x^{2}} $$

Step-by-Step Solution

Answer

The differential of the function \(y=\sqrt{9-x^{2}}\) is \(dy = -\frac{x}{\sqrt{9 - x^2}} dx .\)

Step 1: Recognize the inner function

The inner function here is \(u = 9 - x^2\). Recognize that this function is part of a composition of functions.

Step 2: Differentiate using chain rule

Differentiate the function using the chain rule. The chain rule is a formula to compute the derivative of a composite function. The chain rule states that the derivative of a function composition is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. The derivative of \(y = \sqrt{u}\) is \(\frac{1}{2\sqrt{u}}\) and the derivative of \(u = 9 - x^2\) is \(-2x\). By the chain rule, the derivative of \(y = \sqrt{9 - x^2}\) is \(\frac{1}{2\sqrt{9 - x^2}} \cdot -2x = -\frac{x}{\sqrt{9 - x^2}}\).

Step 3: Compute the differential

The differential of the function \(y\), denoted \(dy\), is defined as the derivative of \(y\) with respect to \(x\), \(y'\), times \(dx\), that is, \(dy = y' dx\). Since we know that \(y'\) = -\(\frac{x}{\sqrt{9 - x^2}}\), then \(dy = -\frac{x}{\sqrt{9 - x^2}} dx .\)

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in differential calculus that helps us find the derivative of composite functions. In simple terms, it allows us to differentiate complex functions by breaking them down into simpler parts.

• When we encounter a composite function, look at the innermost function—this becomes your 'inner function'.
• The outer function will be what remains when the inner function's effects are applied.
• The Chain Rule says that you differentiate the outer function first and then multiply it by the derivative of the inner function.

This process helps untangle complex layers of functions. For example, in the given function, after identifying the inner function as \(u = 9 - x^2\), we find the derivative of the outer function \(y = \sqrt{u}\) and multiply it by the derivative of the inner function \(u\).
This gives us the complete derivative needed to find the differential.

Composite Function

A Composite Function is formed when one function is nested inside another. This means the output of one function becomes the input of the other.

• For instance, if you have two functions, \(f(x)\) and \(g(x)\), the composite function could be \(f(g(x))\).
• It's like doing one process, and then immediately applying a second process on the result of the first.
• In our exercise, \(y = \sqrt{9 - x^2}\) is a composite function. Here, \(9 - x^2\) is the inner function, and applying the square root is the outer function.

Understanding composite functions is key in calculus because it allows us to handle and simplify complex expressions using the chain rule effectively. Once decomposed into these two layers, each function can be tackled individually which simplifies the differentiation process.

Derivatives

Derivatives measure how a function changes as its input changes. They are at the core of differential calculus, allowing us to understand rates of change and slopes of curves.

• The derivative is often denoted as \(f'(x)\), \(\frac{dy}{dx}\), or simply \(y'\).
• The act of differentiating a function involves finding its derivative concerning its input variable, usually \(x\).
• Finding derivatives uses rules like the power rule, product rule, quotient rule, and importantly, the chain rule.

In the exercise, we differentiated the square root function using the chain rule to handle its composite nature. By identifying the derivative of both the inner and outer functions, we can accurately describe the rate at which \(y\) changes with \(x\). Derivatives are incredibly powerful for understanding the behavior of functions, especially when plotting them or determining maxima and minima.

Differential

The Differential, denoted as \(dy\), represents an infinitesimally small change in the function's output, corresponding to a small change \(dx\) in the input.

• It is closely related to the derivative, as \(dy = f'(x) \, dx\).
• The concept emphasizes the change and serves well in approximations.
• In calculus problems, the differential helps in solving problems of rate of change and constructing slopes of tangent lines.

In the given problem, after finding the derivative \(-\frac{x}{\sqrt{9 - x^2}}\), the differential \(dy\) becomes \(-\frac{x}{\sqrt{9 - x^2}} \, dx\). This expresses how a tiny increase in \(x\) leads to a change in \(y\), serving as an essential tool for approximation and analysis in calculus.