Chapter 3: Problem 22
$$ \text { } , \text { find the indicated derivative. } $$ $$ \frac{d r}{d x} \text { if } r=\frac{\ln x}{x^{2} \ln x^{2}}+\left(\ln \frac{1}{x}\right)^{3} $$
Short Answer
Expert verified
\( \frac{dr}{dx} = -\frac{1}{x^3} - \frac{3(\ln x)^2}{x} \)
Step by step solution
01
Simplify the Expression
First, simplify the expression for \( r \). The expression \( r = \frac{\ln x}{x^{2} \ln x^{2}} + \left(\ln \frac{1}{x}\right)^{3} \) can be rewritten by simplifying each part. Notice that \( \ln x^2 = 2 \ln x \). Hence, the first term is \( \frac{\ln x}{x^2 \cdot 2 \ln x} = \frac{1}{2x^2} \). The second term becomes \((\ln x^{-1})^3 = (-\ln x)^3 = - (\ln x)^3 \). So, \( r = \frac{1}{2x^2} - (\ln x)^3 \).
02
Derive the Simplified Function
Now find the derivative \( \frac{dr}{dx} \) of the simplified expression \( r = \frac{1}{2x^2} - (\ln x)^3 \). Use the power and chain rules here. For \( \frac{1}{2x^2} \), rewrite it as \( \frac{1}{2} x^{-2} \). Its derivative is \( -\frac{2}{2}x^{-3} = -\frac{1}{x^3} \). For \( (\ln x)^3 \), use the chain rule: the derivative is \( 3(\ln x)^2 \cdot \frac{1}{x} \).
03
Combine the Derivatives
Combine the derivatives of the two terms. The derivative of \( r \) is \( \frac{dr}{dx} = -\frac{1}{x^3} - \frac{3(\ln x)^2}{x} \). Simplify this to \[ \frac{dr}{dx} = -\frac{1}{x^3} - \frac{3(\ln x)^2}{x} \].
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 Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithmic Differentiation
Logarithmic differentiation is a clever technique used when finding derivatives of functions that involve logarithmic expressions which are difficult to differentiate using standard rules. This can include expressions where the variable appears both in the base and the exponent, or products, and quotients of functions that can be simplified using logarithms.
To apply logarithmic differentiation:
To apply logarithmic differentiation:
- Take the natural logarithm of both sides of the equation.
- Differentiation of both sides using the chain rule.
- Simplify and solve for the derivative.
Chain Rule
The chain rule is a fundamental tool in calculus, essential when differentiating composite functions. It dictates that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
To understand the chain rule, consider a function \(f(g(x))\), where \(f\) is the outer function and \(g\) is the inner one:
To understand the chain rule, consider a function \(f(g(x))\), where \(f\) is the outer function and \(g\) is the inner one:
- Differentiate the outer function while keeping the inner function unchanged.
- Multiply by the derivative of the inner function.
Power Rule
The power rule is often one of the first derivative rules learned and it provides a straightforward method for differentiating functions of the form \(x^n\). The rule is quite simple: the derivative of \(x^n\) is \(nx^{n-1}\). This rule is immensely helpful in various calculus problems, including those which involve simplifying complex expressions.
In the provided exercise, the power rule is applied to terms that have expressions in the form of \(x\) raised to a power. For example:
In the provided exercise, the power rule is applied to terms that have expressions in the form of \(x\) raised to a power. For example:
- To derive \(\frac{1}{2x^2}\), first express it as \( \frac{1}{2} x^{-2} \). The derivative is calculated as \( -2 \cdot \frac{1}{2} x^{-3} = -\frac{1}{x^3} \).