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In Exercises 1 through \(34,\) find the derivative. $$ f(x)=\frac{x^{2}+1}{\left(x^{4}+x+1\right)^{4}} $$

Short Answer

Expert verified
Use the Quotient Rule and Chain Rule to differentiate.

Step by step solution

01

Identify the Rule to Use

The function \( f(x) = \frac{x^2 + 1}{(x^4 + x + 1)^4} \) is a rational function, i.e., a quotient of two functions. To find its derivative, we should use the Quotient Rule, which is: \( \frac{u}{v}' = \frac{u'v - uv'}{v^2} \), where \( u = x^2 + 1 \) and \( v = (x^4 + x + 1)^4 \).
02

Differentiate the Numerator

Let \( u = x^2 + 1 \). The derivative of \( u \) with respect to \( x \) is found by differentiating term by term: \( u' = \frac{d}{dx}(x^2) + \frac{d}{dx}(1) = 2x + 0 = 2x \).
03

Differentiate the Denominator

Let \( v = (x^4 + x + 1)^4 \). To find \( v' \), we apply the Chain Rule. Set \( g(x) = x^4 + x + 1 \) so \( v = g(x)^4 \). The Chain Rule gives us \( v' = 4 \cdot g(x)^3 \cdot g'(x) \). First, calculate \( g'(x) = 4x^3 + 1 \). Thus, \( v' = 4(x^4 + x + 1)^3(4x^3 + 1) \).
04

Apply the Quotient Rule

Using the Quotient Rule: \( \frac{u}{v}' = \frac{u'v - uv'}{v^2} \), substitute \( u = x^2 + 1 \), \( u' = 2x \), \( v = (x^4 + x + 1)^4 \), and \( v' = 4(x^4 + x + 1)^3(4x^3 + 1) \). This gives:\[f'(x) = \frac{(2x)((x^4 + x + 1)^4) - (x^2 + 1)(4(x^4 + x + 1)^3(4x^3 + 1))}{((x^4 + x + 1)^4)^2}\]
05

Simplify the Expression

Simplify the expression for \( f'(x) \):\[f'(x) = \frac{2x(x^4 + x + 1)^4 - 4(x^2 + 1)(x^4 + x + 1)^3(4x^3 + 1)}{(x^4 + x + 1)^8}\].As this rational expression could be simplified further by factoring, seek for any common terms that might cancel or further computations that clarify the numerator.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
When dealing with derivatives of rational functions, the quotient rule is a powerful tool. The rule applies whenever you have a function that can be expressed as a ratio, or quotient, of two functions. Given a function expressed as \( f(x) = \frac{u}{v} \), the quotient rule helps us find the derivative \( f'(x) \) using the formula:\[ f'(x) = \frac{u'v - uv'}{v^2} \]
  • \( u \) is the numerator function, and \( u' \) represents its derivative.
  • \( v \) is the denominator function, and \( v' \) is its derivative.
In the example function \( f(x) = \frac{x^2 + 1}{(x^4 + x + 1)^4} \), we identify:
  • \( u = x^2 + 1 \), and its derivative \( u' = 2x \).
  • \( v = (x^4 + x + 1)^4 \), for which we will need to employ other techniques to find \( v' \).
Using this rule helps simplify and correctly calculate the derivative of such complex fractions, ensuring nothing is missed in the computation process.
Chain Rule
The chain rule is essential when differentiating composite functions. A composite function is something like \( (g(x))^n \), where one function is nested inside another. The rule states that to differentiate \( (g(x))^n \), you multiply the derivative of the outer function by the derivative of the inner function.To differentiate \( v = (x^4 + x + 1)^4 \), identify its inner function \( g(x) = x^4 + x + 1 \). Thus, \( v = g(x)^4 \). The chain rule then tells us:\[ v' = 4 \cdot (g(x))^3 \cdot g'(x) \]
  • The outer function is \( (g(x))^4 \), whose derivative is \( 4 \cdot (g(x))^3 \).
  • The derivative of \( g(x) = x^4 + x + 1 \) is \( g'(x) = 4x^3 + 1 \).
Substituting these into the formula gives:\[ v' = 4(x^4 + x + 1)^3(4x^3 + 1) \]This technique allows us to break down and manage function layers effectively, making difficult differentiations approachable.
Differentiation Techniques
Differentiation is a cornerstone of calculus, used for finding the rate at which a function is changing at any given point. There are various techniques, with each suited for different types of functions.
  • Power Rule: A simple rule used for monomials, expressed as \( \frac{d}{dx}(x^n) = nx^{n-1} \).
  • Product Rule: Useful for differentiating products of two functions, stating \( (uv)' = u'v + uv' \).
  • Quotient Rule: Handy for rational functions, as discussed earlier.
  • Chain Rule: For composite functions, already explored in detail.
These techniques often work together, as in this exercise, where both the quotient and chain rules are utilized. Knowing when and how to apply each is key to mastering calculus. Practicing these rules on various functions helps build intuition and confidence.

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