Problem 27 - Thomas Calculus

Question

Find the derivatives of the functions $$ y=\frac{1}{\left(x^{2}-1\right)\left(x^{2}+x+1\right)} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{-4x^3 - 3x^2 + 1}{(x^2-1)^2 (x^2+x+1)^2} \).

Step 1: Identify the Derivative Rule

The given function is a rational function in the form of a quotient. We'll need to apply the Quotient Rule of differentiation, which is \( \left(\frac{u}{v}\right)' = \frac{v u' - u v'}{v^2} \), where \( u \) is the numerator and \( v \) is the denominator of the function.

Step 2: Differentiate the Numerator

In this case, the numerator \( u = 1 \), which is a constant. The derivative of a constant is zero, so \( u' = 0 \).

Step 3: Differentiate the Denominator

The denominator \( v = (x^2-1)(x^2+x+1) \) is a product of two functions. Use the Product Rule \( (fg)' = f'g + fg' \) to find \( v' \).

Step 4: Apply the Product Rule to Differentiate v

Let \( f = x^2 - 1 \) and \( g = x^2 + x + 1 \). Then \( f' = 2x \) and \( g' = 2x + 1 \). The product rule gives: \[ v' = (2x)(x^2 + x + 1) + (x^2 - 1)(2x + 1). \]

Step 5: Simplify the Expression for v'

Expand \[(2x)(x^2 + x + 1) = 2x^3 + 2x^2 + 2x\] and \[(x^2 - 1)(2x + 1) = 2x^3 + x^2 - 2x - 1.\]Combine the expressions to get \[v' = 4x^3 + 3x^2 + 0x - 1 = 4x^3 + 3x^2 - 1.\]

Step 6: Apply the Quotient Rule

Using the Quotient Rule formula:\[ \left(\frac{1}{(x^2-1)(x^2+x+1)}\right)' = \frac{(x^2-1)(x^2+x+1) \cdot 0 - 1 \cdot (4x^3 + 3x^2 - 1)}{((x^2-1)(x^2+x+1))^2}. \]

Step 7: Simplify the Derivative

The first term in the numerator is zero, so \[-(4x^3 + 3x^2 - 1)\] becomes \[-4x^3 - 3x^2 + 1.\]Therefore, the derivative is \[\frac{-4x^3 - 3x^2 + 1}{(x^2-1)^2 (x^2+x+1)^2}.\]

Key Concepts

Quotient Rule

When differentiating certain types of functions, especially rational functions, the quotient rule becomes an essential tool. A rational function is essentially a function divided by another function. In these cases, the derivative can be found using the quotient rule. This rule comes in handy when you have a function represented as a fraction, termed as \( \frac{u}{v} \). In the quotient rule, \( u \) is the numerator and \( v \) is the denominator. The quotient rule formula is:\[ \left(\frac{u}{v}\right)' = \frac{vu' - uv'}{v^2} \]One of the main tasks is to differentiate both the numerator \( u \) and the denominator \( v \). Once the derivatives \( u' \) and \( v' \) are calculated, they must be appropriately combined according to the formula. Don’t forget to square the denominator to complete the rule.A key thing to remember: always be careful with algebraic simplification. Each part of the quotient rule relies on your ability to precisely simplify expressions for easy computation.

Product Rule

Another vital rule for differentiation is the product rule. You may have multiple functions that you need to distinguish individually, but appear as products. This is crucial when your function's denominator consists of two or more multiplied expressions.The product rule states:\[ (fg)' = f'g + fg' \]In practical terms, if you've got two functions multiplied, like \( f(x) \) and \( g(x) \), you'll need to:

• Derrickiate one function, leaving the other unchanged.
• Switch, and differentiate the other one.
• Multiply the results based on the rule.

For example, if you have \( f(x) = x^2 - 1 \) and \( g(x) = x^2 + x + 1 \), their derivatives are \( f'(x) = 2x \) and \( g'(x) = 2x + 1 \). Plug these into the product rule: \[ (2x)(x^2 + x + 1) + (x^2 - 1)(2x + 1) \]Simplifying each part separately will make the final derivative assembly easier.

Rational Functions

Rational functions play an important role in calculus, representing the core case for using both the quotient and product rules. A rational function is represented as the ratio of two polynomial functions.In calculus, you often work with rational functions to analyze rates of change, optimize equations, or even solve real-world problems. Understanding their structures is key because these functions can get complex:

• The numerator and the denominator can sometimes be simplified further using algebra.
• Graphically, rational functions can harbor asymptotes, both vertical and horizontal, which impact the function's behavior.

These can be simplified using common factor polynomials or just carefully applying the differentiation rules. Studying how these expressions change as different values of \( x \) are plugged in helps provide intuition beyond mechanical computation. Seeking out these core components will strengthen your differentiation skills for tackling future calculus problems.