Chapter 4: Problem 2
Find the derivative of each function. $$ f(x)=\frac{\ln x}{x^{3}} $$
Short Answer
Expert verified
The derivative is \( f'(x) = \frac{1 - 3 \ln x}{x^4} \).
Step by step solution
01
Identify the function type
The function given is \( f(x) = \frac{\ln x}{x^3} \). This is a quotient of two functions, where the numerator is \( u(x) = \ln x \) and the denominator is \( v(x) = x^3 \).
02
Apply the Quotient Rule
The quotient rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their quotient \( \frac{u}{v} \) is given by \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \].
03
Differentiate the Numerator
The derivative of the numerator \( u(x) = \ln x \) is \( u'(x) = \frac{1}{x} \).
04
Differentiate the Denominator
The derivative of the denominator \( v(x) = x^3 \) is \( v'(x) = 3x^2 \).
05
Plug into the Quotient Rule Formula
Substitute \( u(x), u'(x), v(x), v'(x) \) into the quotient rule: \[ \left( \frac{\ln x}{x^3} \right)' = \frac{\left(\frac{1}{x}\right)x^3 - (\ln x)3x^2}{(x^3)^2} \].
06
Simplify the Expression
Simplify the expression obtained from the quotient rule: 1. \( \frac{x^3}{x} = x^2 \), so the first term is \( x^2 \).2. The second term is \(3x^2 \ln x\).3. The denominator becomes \(x^6\).The expression simplifies to \[ \frac{x^2 - 3x^2 \ln x}{x^6} \].
07
Further Simplify the Fraction
Factor out \(x^2\) from the numerator: \[ \frac{x^2(1 - 3 \ln x)}{x^6} \]. This reduces to \[ \frac{1 - 3 \ln x}{x^4} \] by cancelling \(x^2\) in the numerator and denominator.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
When finding the derivative of a fraction that involves two functions, the quotient rule is your go-to tool. This rule helps us differentiate functions like \( \frac{u}{v} \), where both \( u \) and \( v \) are themselves functions of \( x \). The quotient rule formula is:
\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}.\]
To apply the quotient rule, you'll need to:
\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}.\]
To apply the quotient rule, you'll need to:
- Distantiate the numerator, getting \( u'(x) \).
- Distantiate the denominator, determining \( v'(x) \).
- Plug these derivatives into the formula above.
- Simplify to obtain your final expression.
Logarithmic Functions
Logarithmic functions, such as \( \ln x \), are very common in calculus. These functions have unique properties that come in handy during differentiation. The natural logarithm, denoted by \( \ln x \), has a straightforward derivative:
\[\frac{d}{dx}(\ln x) = \frac{1}{x}.\]
This simple formula is derived from the relationship between exponential and logarithmic functions.
\[\frac{d}{dx}(\ln x) = \frac{1}{x}.\]
This simple formula is derived from the relationship between exponential and logarithmic functions.
Working with \( \ln x \)
The rate of change of \( \ln x \) is slower compared to polynomial functions, which often gives it distinct graph characteristics:- \( \ln x \) is only defined for positive \( x \), so we restrict our domain to \( x > 0 \).
- It increases, albeit slowly, for larger \( x \) values.
- Its derivative, \( \frac{1}{x} \), indicates a decreasing slope as \( x \) increases.
Power Functions
Power functions appear in many calculus problems. They are functions of the form \( x^n \), where \( n \) is a constant. Finding their derivative is straightforward using the power rule:
\[\frac{d}{dx}(x^n) = nx^{n-1}.\]
This rule tells us that you multiply the original power \( n \) by the coefficient in front of \( x \), then decrease the power by one.
\[\frac{d}{dx}(x^n) = nx^{n-1}.\]
This rule tells us that you multiply the original power \( n \) by the coefficient in front of \( x \), then decrease the power by one.
Applying the Power Rule
Here's what you need to know about using the power rule:- It works for any real number \( n \), whether the power is positive, negative, or a fraction.
- Each time you apply this rule, you simplify the expression by reducing the power by one.
- Power functions have derivatives that become constant when \( n = 1 \) because \( \frac{d}{dx}(x) = 1 \).