Problem 28 - Essential Calculus Early Transcendentals

Question

Find the derivative of the function. Simplify where possible. \(y=\arctan \sqrt{\frac{1-x}{1+x}}\)

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -\frac{1}{(\sqrt{\frac{1-x}{1+x}})(1+x)} \).

Step 1: Recognize the Composite Function

The given function is a composite function involving the arctangent function and a square root. Specifically, it is composed as \( y = \arctan(z) \) where \( z = \sqrt{\frac{1-x}{1+x}} \).

Step 2: Differentiate the Arctangent

The derivative of the arctangent function, \( \arctan(z) \), with respect to \( z \) is \( \frac{1}{1+z^2} \). Thus, \( \frac{dy}{dz} = \frac{1}{1+z^2} \).

Step 3: Simplify the Expression Inside Arctangent

Substitute \( z = \sqrt{\frac{1-x}{1+x}} \) into the equation to find \( z^2 = \frac{1-x}{1+x} \). This implies \( 1+z^2 = 1 + \frac{1-x}{1+x} = \frac{1-x+1+x}{1+x} = \frac{2}{1+x} \).

Step 4: Differentiate the Inner Function

Differentiate \( z = \sqrt{\frac{1-x}{1+x}} \) using the chain rule. First find the derivative of \( \frac{1-x}{1+x} \) with respect to \( x \) using the quotient rule: \[ \frac{d}{dx} \left( \frac{1-x}{1+x} \right) = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = -\frac{2}{(1+x)^2}. \]

Step 5: Apply the Chain Rule

Finally, apply the chain rule to differentiate \( z = \sqrt{u} \) (where \( u = \frac{1-x}{1+x} \)): \( \frac{dz}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{\frac{1-x}{1+x}}} \cdot \left(-\frac{2}{(1+x)^2}\right) = -\frac{1}{(1+x)\sqrt{\frac{1-x}{1+x}}} \).

Step 6: Combine Results Using Chain Rule

From Step 2, find \( \frac{dy}{dx} \) using the chain rule: \[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} = \frac{1}{1+z^2} \times -\frac{1}{(1+x)\sqrt{\frac{1-x}{1+x}}} = -\frac{1}{(1+x)\sqrt{\frac{1-x}{1+x}}} \times \frac{1+x}{2} = -\frac{1}{2\sqrt{\frac{1-x}{1+x}}} \cdot \frac{2}{1+x}. \] Simplifying further provides the result for \( \frac{dy}{dx} \).

Step 7: Final Simplification

Since the middle part already simplifies to \( \frac{2}{(1+x)^2} \), oversee the entire chain reaction from Steps 2 to 6 and reorganize it into: \[ \frac{dy}{dx} = -\frac{2}{2(\sqrt{\frac{1-x}{1+x}})(1+x)} = -\frac{1}{(\sqrt{\frac{1-x}{1+x}})(1+x)}, \]which can be simplified further to be a function of \( x \) rather than its possible nested expressions.

Key Concepts

Composite Functions

In calculus, composite functions play a crucial role in understanding more complex expressions. A composite function combines two or more functions into one, where the output of one function becomes the input of another.
For example, in the given problem, the function is represented by the arctangent function applied to a square root. Thus, our main function is written as \( y = \arctan(z) \) with another function defined within it: \( z = \sqrt{\frac{1-x}{1+x}} \).
Recognizing composite functions helps set the stage for differentiation because each layer or function component requires a different approach or rule to find its derivative. By deconstructing the problem into its parts—like separating \( \arctan(z) \) and the square root expression—you can better manage and understand each component's contribution to the whole function.

Chain Rule

The chain rule is a fundamental principle in calculus used when differentiating composite functions. It allows us to find the derivative of compositions by taking the derivative of the outer function and multiplying it by the derivative of the inner function.
Consider our specific exercise: for \( y = \arctan(z) \) where \( z = \sqrt{\frac{1-x}{1+x}} \), the chain rule enables us to combine derivatives to find \( \frac{dy}{dx} \).
First, differentiate the outer function: the derivative of \( \arctan(z) \) in terms of \( z \) is \( \frac{1}{1 + z^2} \).
Second, use the chain rule to handle \( z \). We find \( \frac{dz}{dx} \) by recognizing \( z \) itself is a composition needing its own set of rules—like the quotient and chain rule for differentiating \( \sqrt{u} \) where \( u = \frac{1-x}{1+x} \).
Applying \( \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \) is a quintessential step in managing several layers of nested functions efficiently.

Quotient Rule

The quotient rule is pivotal when dealing with fractions or rational functions during differentiation. This rule is used when a function can be expressed as a ratio of two differentiable functions.
Given two functions \( u(x) \) and \( v(x) \), the quotient rule says: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \).
In our exercise, the expression \( \frac{1-x}{1+x} \) involves such a ratio.

• Differentiate the numerator \( (1-x) \) and denominator \( (1+x) \) separately.
• Plug these derivatives into the quotient rule formula.
• The result shows how these differentiate collectively: \(-\frac{2}{(1+x)^2} \).

This calculated derivative is integral when we apply the chain rule. By breaking up the function into parts, and gradually differentiating, we can handle complex derivatives systematically.