X 2 1 X 2 1 Derivative

Mastering the Derivative of (x²+1)/(x²-1): A Step-by-Step Guide

Calculus opens the door to understanding how things change, and one of its most powerful tools is the derivative. For students and enthusiasts alike, tackling the derivative of a rational function like (x²+1)/(x²-1) presents a perfect opportunity to master a fundamental rule and deepen algebraic intuition. This article will guide you through the entire process, from applying the quotient rule to simplifying the final expression, ensuring you not only compute the answer but also understand the why behind each step. By the end, you’ll have a clear, reusable method for differentiating similar functions and insight into what the resulting derivative actually tells us about the original curve.

Why the Quotient Rule is Your Go-To Tool

Before diving in, it’s crucial to recognize the structure of our function: f(x) = (x²+1)/(x²-1). It is a quotient of two differentiable functions. The numerator, u(x) = x²+1, and the denominator, v(x) = x²-1, are both simple polynomials. When you have a function expressed as one function divided by another, the quotient rule is the mandatory technique. You cannot simply apply the power rule to the entire fraction; that’s a common and critical mistake. The quotient rule states:

If f(x) = u(x)/v(x), then f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]²

Think of it as a "low-d-high minus high-d-low" mnemonic, where "low" is the denominator and "high" is the numerator. This formula elegantly handles the interplay between the rates of change of both the top and bottom parts of the fraction.

Step-by-Step Derivation of (x²+1)/(x²-1)

Let’s apply the rule methodically.

1. Identify u(x), v(x), and their derivatives.

• u(x) = x² + 1 → u'(x) = 2x
• v(x) = x² - 1 → v'(x) = 2x

2. Plug into the quotient rule formula. f'(x) = [ (u'(x) * v(x)) - (u(x) * v'(x)) ] / [v(x)]² f'(x) = [ (2x * (x² - 1)) - ((x² + 1) * 2x) ] / (x² - 1)²

3. Expand the numerator carefully. This is where algebra precision is paramount.

• First term: 2x * (x² - 1) = 2x³ - 2x
• Second term: (x² + 1) * 2x = 2x³ + 2x

Now, substitute these back: Numerator = (2x³ - 2x) - (2x³ + 2x)

4. Simplify the numerator by distributing the negative sign. (2x³ - 2x) - 2x³ - 2x = 2x³ - 2x - 2x³ - 2x

Combine like terms: 2x³ - 2x³ = 0 -2x - 2x = -4x

So, the entire numerator simplifies dramatically to -4x.

5. Write the final, simplified derivative. f'(x) = -4x / (x² - 1)²

This is the elegant, simplified form of the derivative. Notice how the cubic terms canceled out completely. This simplification is not just for neatness; a simpler derivative is easier to analyze for critical points, intervals of increase/decrease, and curve sketching.

The Art of Simplification: Why Did the x³ Terms Vanish?

The cancellation of the 2x³ terms is a beautiful algebraic outcome. It occurs because the derivatives of the numerator and denominator (u'(x) and v'(x)) were identical (both 2x). In the quotient rule structure [u'v - uv'], if u' and v' are proportional or equal in a specific way, significant simplification can happen. Here, u' = v' = 2x, so the terms involving x³ were: (2x * x²) - (x² * 2x) = 2x³ - 2x³ = 0. This pattern is worth remembering: when the derivatives of the numerator and denominator are the same function, the highest-degree terms in the derivative’s numerator will cancel. This insight can save you time and help you check your work.

What Does This Derivative Tell Us? A Graphical Intuition

The derivative f'(x) = -4x / (x² - 1)² is more than a symbolic answer; it’s a new function describing the slope of the original function f(x) = (x²+1)/(x²-1) at any point x.

• The Denominator: (x² - 1)² is always positive for all x ≠ ±1 (where the original function is undefined, having vertical asymptotes). A positive denominator means the sign of f'(x) is determined entirely by the numerator, -4x.
• Sign Analysis:
  • For x > 0, -4x is negative → f'(x) < 0 → f(x) is decreasing.
  • For x < 0, -4x is positive → f'(x) > 0 → f(x) is increasing.
  • At x = 0, f