Chapter 3: Problem 11
Find the derivative. $$ f(x)=\frac{x}{\left(x^{2}-1\right)^{4}} $$
Short Answer
Expert verified
The derivative is \( f'(x) = \frac{-7x^2}{(x^2 - 1)^5} \).
Step by step solution
01
Recognize the Form
The function given is a quotient of two functions, specifically \( f(x) = \frac{u(x)}{v(x)} \), where \( u(x) = x \) and \( v(x) = (x^2 - 1)^4 \). We should apply the quotient rule to differentiate the function.
02
Differentiate the Numerator
Find the derivative of the numerator \( u(x) = x \). The derivative is straightforward: \( u'(x) = 1 \).
03
Differentiate the Denominator
Now differentiate the denominator \( v(x) = (x^2 - 1)^4 \). Use the chain rule. Start by letting \( g(x) = x^2 - 1 \), then \( v(x) = [g(x)]^4 \). The derivative is \( v'(x) = 4[g(x)]^3 \, g'(x) \), where \( g'(x) = 2x \). Thus, \( v'(x) = 4(x^2 - 1)^3 \, (2x) = 8x(x^2 - 1)^3 \).
04
Apply the Quotient Rule
The quotient rule formula for derivatives is \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Substitute the derived components: \( u'(x) = 1 \), \( u(x) = x \), \( v(x) = (x^2 - 1)^4 \), and \( v'(x) = 8x(x^2 - 1)^3 \).
05
Simplify the Expression
Plug in to find: \( f'(x) = \frac{1 \cdot (x^2 - 1)^4 - x \cdot 8x(x^2 - 1)^3}{((x^2 - 1)^4)^2} \). Simplifying the numerator, we have \( (x^2 - 1)^4 - 8x^2(x^2 - 1)^3 \). Factoring out \( (x^2 - 1)^3 \) gives \( (x^2 - 1)^3[(x^2 - 1) - 8x^2] \). The expression becomes \( -7x^2(x^2 - 1)^3 \).
06
Final Form of the Derivative
The complete expression for the derivative is: \[ f'(x) = \frac{-7x^2(x^2 - 1)^3}{(x^2 - 1)^8} \]. Simplifying, we can cancel \((x^2 - 1)^3\) from the numerator and the denominator. The final result is \[ f'(x) = \frac{-7x^2}{(x^2 - 1)^5} \].
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
When faced with a function that is a division of two smaller functions, you need a special tool for differentiation: the quotient rule. This is pivotal when you encounter a function like \( f(x) = \frac{u(x)}{v(x)} \). The quotient rule allows us to differentiate such functions systematically. The rule is expressed as:
- \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)
Chain Rule
The chain rule is essential when you deal with composite functions, functions within functions. In our context, it neatly handles situations like when you have \( v(x) = (x^2 - 1)^4 \) since it involves a function \( g(x) = x^2 - 1 \) raised to a power. The essence of the chain rule is:
- If \( h(x) = f(g(x)) \), then its derivative \( h'(x) = f'(g(x)) \cdot g'(x) \).
- Differentiate the outer function while keeping the inner function untouched.
- Then, multiply this derivative by the derivative of the inner function.
Differentiation
Differentiation is the mathematical process of finding a function's derivative, or its rate of change. It uncovers how the function's value changes as its input changes, a key concept in calculus. For any function \( f(x) \):
- The derivative \( f'(x) \) provides a powerful tool for understanding the curve's slope at any point \( x \).
- the differentiation process applied the quotient rule to a function expressed as a fraction, \( \frac{x}{(x^2 - 1)^4} \).
- Also, it used the chain rule to handle the inner composition of functions in the denominator.