Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 3: Problem 18
Find the derivative of the function. \(f(x)=\frac{\left(x^{2}+1\right)^{2}}{\left(x^{4}+1\right)^{4}}\)
Short Answer
Expert verified
The derivative is \( f'(x) = \frac{4x(x^2 + 1)(x^4 + 1)^3(-3x^4 - 4x^2 + 1)}{(x^4 + 1)^8} \).
Step by step solution
01
Identify the quotient rule
The given function is a fraction of the form \( \frac{u(x)}{v(x)} \) where \( u(x) = (x^2 + 1)^2 \) and \( v(x) = (x^4 + 1)^4 \). To find the derivative \( f'(x) \), we will apply the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \).
02
Find the derivative of the numerator
The numerator of the function is \( u(x) = (x^2 + 1)^2 \). We use the chain rule for differentiation: \( u'(x) = 2(x^2 + 1) \cdot (2x) = 4x(x^2 + 1) \).
03
Find the derivative of the denominator
The denominator of the function is \( v(x) = (x^4 + 1)^4 \). Again using the chain rule: \( v'(x) = 4(x^4 + 1)^3 \cdot (4x^3) = 16x^3(x^4 + 1)^3 \).
04
Apply the quotient rule
Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the quotient rule formula:\[f'(x) = \frac{(x^4 + 1)^4 \cdot 4x(x^2 + 1) - (x^2 + 1)^2 \cdot 16x^3(x^4 + 1)^3}{((x^4 + 1)^4)^2}\]
05
Simplify the expression
Factor out common terms from the numerator: \[f'(x) = \frac{4x(x^2 + 1)(x^4 + 1)^3 \left((x^4 + 1) - 4x^2(x^2 + 1)\right)}{(x^4 + 1)^8}\]Further simplify:\[(x^4 + 1) - 4x^2(x^2 + 1) = x^4 + 1 - 4x^4 - 4x^2 = -3x^4 - 4x^2 + 1\]
06
Write the final simplified derivative
The final derivative is:\[f'(x) = \frac{4x(x^2 + 1)(x^4 + 1)^3(-3x^4 - 4x^2 + 1)}{(x^4 + 1)^8}\]This can be further simplified, but for clarity, this is the expression of \( f'(x) \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
When differentiating a function that is a fraction, the quotient rule is an essential tool. It is used when you have a function in the form \( \frac{u(x)}{v(x)} \), where both the numerator \( u(x) \) and the denominator \( v(x) \) are functions of the same variable. This rule helps you find the derivative by applying the formula: \[ \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \]
- First, find the derivatives of the numerator \( u'(x) \) and the denominator \( v'(x) \).
- Next, multiply the denominator function \( v(x) \) by the derivative of the numerator \( u'(x) \).
- Then, subtract the result of the numerator \( u(x) \) times the derivative of the denominator \( v'(x) \).
- Finally, divide the entire expression by the square of the denominator function \( v(x)^2 \).
Chain Rule
Differentiating functions that are nested within each other requires the chain rule. It is particularly useful when dealing with composite functions, allowing you to find the derivative of a function that is, literally, a function within a function. For a composite function \( Y = f(g(x)) \), the derivative \( Y' \) can be found using the formula: \[ Y' = f'(g(x)) \cdot g'(x) \]
- First, identify the outer function \( f \) and the inner function \( g \).
- Take the derivative of the outer function, \( f'(g(x)) \).
- Then, multiply it by the derivative of the inner function, \( g'(x) \).
Simplification
After finding the derivative using the quotient and chain rules, you often face a complex expression. Simplification is the process of reducing this expression to its simplest or most intuitive form. Here is how you can approach it:
- Factor out common terms, which can reduce the complexity of the expression.
- Combine like terms, which simplifies the arithmetic operations involved.
- Break down expressions, sometimes by rewriting and reorganizing them with basic algebraic identities.
