Problem 38 - Calculus Early Transcendentals

Question

Find \(d y / d x\) $$ y=\left(\frac{1+x^{2}}{1-x^{2}}\right)^{17} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 17 \left( \frac{1+x^2}{1-x^2} \right)^{16} \cdot \frac{4x}{(1-x^2)^2} \)

Step 1: Identify the Function Type

The function given is a composite function raised to a power. Specifically, it's a rational function \( \left( \frac{1+x^{2}}{1-x^{2}} \right) \) raised to the 17th power.

Step 2: Apply the Chain Rule

To differentiate this function, use the chain rule. First, define \( u = \frac{1+x^{2}}{1-x^{2}} \). Then \( y = u^{17} \). The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).

Step 3: Differentiate the Outer Function

Differentiating the outer function \( y = u^{17} \) with respect to \( u \) gives \( \frac{dy}{du} = 17u^{16} \).

Step 4: Differentiate the Inner Function

Find \( \frac{du}{dx} \) using the quotient rule, where \( u = \frac{1+x^{2}}{1-x^{2}} \). The quotient rule is \( \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \). Set \( f(x) = 1 + x^2 \) and \( g(x) = 1 - x^2 \).

Step 5: Apply the Quotient Rule

Calculate \( f'(x) = 2x \) and \( g'(x) = -2x \). Then apply the quotient rule: \[ \frac{du}{dx} = \frac{(2x)(1-x^2) - (1+x^2)(-2x)}{(1-x^2)^2} = \frac{2x - 2x^3 + 2x + 2x^3}{(1-x^2)^2}. \] Simplifying, we get: \[ \frac{du}{dx} = \frac{4x}{(1-x^2)^2}. \]

Step 6: Combine Results from Chain Rule

Substitute \( u = \frac{1+x^2}{1-x^2} \), \( \frac{dy}{du} = 17u^{16} \), and \( \frac{du}{dx} = \frac{4x}{(1-x^2)^2} \) into the chain rule: \[ \frac{dy}{dx} = 17 \left( \frac{1+x^2}{1-x^2} \right)^{16} \cdot \frac{4x}{(1-x^2)^2}. \]

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental technique in calculus differentiation, used when we need to take the derivative of a composition of functions. In simpler terms, it helps us differentiate a function that is nested within another function.
For instance, in the given problem, you have a function that has been raised to the power of 17. Inside this power function, there's another function, which is a fraction. To differentiate such a composite setup, the chain rule guides us.
The formula for the chain rule is expressed as:

• \( rac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \),

where \( u \) is the inside function. In our exercise, defining the inside function as \( u = \frac{1+x^2}{1-x^2} \) makes the outer function \( y = u^{17} \). The chain rule helps us handle these layers step by step, ensuring each component is correctly differentiated.

The Quotient Rule Simplified

When dealing with fractions in calculus, the quotient rule is an essential tool for differentiation. If you have a function expressed as a quotient, such as \( \frac{f(x)}{g(x)} \), the quotient rule will guide you in finding its derivative.
The formula is:

• \( \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \).

Here, \( f(x) \) and \( g(x) \) are the numerator and the denominator, respectively.
In our task, the quotient rule was applied to \( u = \frac{1+x^2}{1-x^2} \). By setting \( f(x) = 1 + x^2 \) and \( g(x) = 1 - x^2 \), we utilized this rule to handle their derivatives, \( f'(x) \) being \( 2x \) and \( g'(x) \) being \( -2x \). This application is crucial in finding \( \frac{du}{dx} \), ultimately contributing to solving the chain rule.

Demystifying Composite Functions

Composite functions consist of two or more functions nested together. In these cases, one function's output becomes the input of another function. They are crucial in calculus as they represent real-world phenomena where decisions or changes are layered.
When you have a function like \( y = \left(\frac{1+x^2}{1-x^2}\right)^{17} \), it's a classic example of a composite function. Not only do you have the power of 17 affecting the whole, but within, there lies a ratio of two functions to contend with.
To successfully differentiate such composite functions, understanding and applying the chain rule and quotient rule becomes imperative. Recognizing how one function nests inside another allows for breaking down the problem and tackling each part separately for easier derivative computation.

Mastering Derivative Computation

Finding the derivative, particularly in complex functions, emerges from mastering the differential rules like the chain rule and the quotient rule. Let's outline the steps:

• Identify the structure, whether it's a simple, product, quotient, or composite function.
• Apply the right technique. In composites, use the chain rule; for fractions, employ the quotient rule.
• Calculate derivatives of individual parts first—here, the outer \( u^{17} \) and the inner \( \frac{1+x^2}{1-x^2} \).

In our example, once these rules are set and used effectively, combining the results provides the overall derivative, \( \frac{dy}{dx} = 17 \left( \frac{1+x^2}{1-x^2} \right)^{16} \cdot \frac{4x}{(1-x^2)^2} \). This meticulous approach will not only bring clarity but also precision in tackling even the most tangled functions.