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Chapter 6: Problem 328

For the following exercises, compute dy/dx by differentiating ln y. $$y=x^{-1 / \ln x}$$

Short Answer

Expert verified
The derivative \(\frac{dy}{dx} = 0\).

Step by step solution

01

Apply the Natural Logarithm

To solve for \(\frac{dy}{dx}\), start by applying the natural logarithm to both sides of the equation. Given \(y = x^{-1/\ln{x}}\), take the natural logarithm:\[\ln{y} = \ln{(x^{-1/\ln{x}})}\] This simplifies to:\[\ln{y} = -\frac{1}{\ln{x}} \cdot \ln{x} = -1\]
02

Differentiate Both Sides

Now, differentiate both sides of the equation with respect to \(x\). Differentiating the left side:\[\frac{d}{dx} (\ln{y}) = \frac{1}{y} \cdot \frac{dy}{dx}\]The derivative of the right side (which is a constant \(-1\)) with respect to \(x\):\[\frac{d}{dx}(-1) = 0\]Thus, we have:\[\frac{1}{y} \cdot \frac{dy}{dx} = 0\]
03

Solve for \(\frac{dy}{dx}\)

Solve the equation \(\frac{1}{y} \cdot \frac{dy}{dx} = 0\) for \(\frac{dy}{dx}\) by multiplying both sides by \(y\):\[\frac{dy}{dx} = y \cdot 0 = 0\]Hence, \(\frac{dy}{dx} = 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of Logarithmic Functions
Derivatives are fundamental in calculus as they measure how a function changes as its input changes. For logarithmic functions, the key rule is essential. The derivative of the natural logarithm, \( \ln(x) \), is \( \frac{1}{x} \). Applying this rule helps in differentiating more complex expressions involving logs.

In the given exercise, we have a function in terms of \( y \) defined as \( y = x^{-1/\ln{x}} \). The first step to find \( \frac{dy}{dx} \) is to apply a logarithm on both sides. This allows us to simplify and facilitate the differentiation process:

  • Taking logarithms helps transform multiplicative and exponential expressions into more manageable additive forms.
  • For differentiating the expression \( y = x^{-1/\ln{x}} \), finding the derivative of \( \ln(y) \) simplifies the process.
Once the logarithm is applied, we then differentiate using the chain and power rules where necessary.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). It's an important function in calculus due to its unique properties:

  • The derivative of \( \ln(x) \) is always \( \frac{1}{x} \), making it handy to simplify complex differentiation.
  • It turns products into sums, which is particularly useful when dealing with exponents and roots.
  • In our exercise, understanding \( \ln{x} \)'s behavior is crucial for simplifying the expression \( x^{-1/\ln{x}} \).
By applying the natural logarithm to both sides of the equation, it converts tricky expressions into linear forms, which are simpler to differentiate.

The expression \( \ln{(x^{-1/\ln{x}})} = -\frac{1}{\ln{x}} \cdot \ln{x} \) ultimately simplifies to a constant \(-1\), making the differentiation straightforward.
Solving Differential Equations
Solving differential equations involves finding an expression for the derivative \( \frac{dy}{dx} \). In the context of this exercise, we set up the equation \( \frac{1}{y} \cdot \frac{dy}{dx} = 0 \) after differentiating both sides.

To solve this:
  • Since the right-hand side of the equation is zero, it implies that the entire left-hand side must equal zero.
  • When solving \( \frac{1}{y} \cdot \frac{dy}{dx} = 0 \), we multiply through by \( y \), resulting in \( \frac{dy}{dx} = 0 \).
This indicates that the rate of change of \( y \) with respect to \( x \) is zero, meaning \( y \) is constant as \( x \) changes.

Understanding how to manipulate the equation to reach this conclusion is a key part of solving differential equations, showcasing the synergy between algebraic manipulation and calculus principles.

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For the following exercises, find the work done. A \(1-m\) spring requires 10 \(\mathrm{J}\) to stretch the spring to 1.1 \(\mathrm{m}\) . How much work would it take to stretch the spring from 1 \(\mathrm{m}\) to 1.2 \(\mathrm{m}\) ?

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