Chapter 4: Problem 26
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.
- \( u = x^2 + 1 \), and its derivative \( u' = 2x \).
- \( v = (x^4 + x + 1)^4 \), for which we will need to employ other techniques to find \( v' \).
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 \( g'(x) = 4x^3 + 1 \).
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.