Chapter 1: Problem 30
For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$ y=x^{-1 / x} $$
Short Answer
Expert verified
\( \frac{dy}{dx} = x^{-1/x} \left( \frac{\ln x - 1}{x^2} \right) \).
Step by step solution
01
Take the Logarithm
Start by taking the natural logarithm of both sides of the equation: \( y = x^{-1/x} \). This gives us \( \ln y = \ln(x^{-1/x}) \). Simplify the right side to get \( \ln y = -\frac{1}{x} \ln x \).
02
Differentiate Both Sides
Differentiate both sides with respect to \(x\). The left side becomes \( \frac{1}{y} \frac{dy}{dx} \) by the chain rule, and the right side requires the product rule: \[ \frac{d}{dx} \left(-\frac{1}{x} \ln x \right) = \frac{d}{dx}(-\frac{1}{x}) \cdot \ln x + (-\frac{1}{x}) \cdot \frac{d}{dx}(\ln x) \].Calculating these, we get \( \frac{d}{dx}(-\frac{1}{x}) = \frac{1}{x^2} \) and \( \frac{d}{dx}(\ln x) = \frac{1}{x} \). Substitute back to find:\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x^2} \ln x - \frac{1}{x^2} \].
03
Solve for \( \frac{dy}{dx} \)
Multiply both sides by \( y \) to solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = y \left( \frac{1}{x^2} \ln x - \frac{1}{x^2} \right) \].Substitute \( y = x^{-1/x} \) back into the equation:\[ \frac{dy}{dx} = x^{-1/x} \left( \frac{1}{x^2} \ln x - \frac{1}{x^2} \right) \].Simplify to get the final result:\[ \frac{dy}{dx} = x^{-1/x} \left( \frac{\ln x - 1}{x^2} \right) \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithmic Differentiation
Logarithmic differentiation is a highly useful technique, especially when dealing with complex functions like products, quotients, or powers that include variables. The core idea is to take the natural logarithm of both sides of an equation, which helps simplify the differentiation process.
We start by transforming a multiplicative function into an additive one using logarithms, as logarithms turn products into sums and powers into multiples, vastly simplifying differentiation.
For instance, in our exercise where \( y = x^{-1/x} \), applying the logarithm yields \( \ln y = \ln (x^{-1/x}) \). This step simplifies further to \( \ln y = -\frac{1}{x} \ln x \), making the function much easier to handle when differentiating.
Logarithmic differentiation effectively handles these complex forms and is particularly handy with functions that are products or powers.
We start by transforming a multiplicative function into an additive one using logarithms, as logarithms turn products into sums and powers into multiples, vastly simplifying differentiation.
For instance, in our exercise where \( y = x^{-1/x} \), applying the logarithm yields \( \ln y = \ln (x^{-1/x}) \). This step simplifies further to \( \ln y = -\frac{1}{x} \ln x \), making the function much easier to handle when differentiating.
Logarithmic differentiation effectively handles these complex forms and is particularly handy with functions that are products or powers.
Chain Rule
The chain rule is fundamental in calculus, allowing us to differentiate composite functions. Imagine a function that consists of one function nested inside another, such as \( f(g(x)) \).
The chain rule formula is \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \). This means you differentiate the outer function, evaluated at the inner function, and then multiply it by the derivative of the inner function.
In the current exercise, we apply the chain rule to the left side of our equation after taking the logarithm, \( \ln y \). It allows us to express \( \frac{d}{dx}(\ln y) \) as \( \frac{1}{y} \frac{dy}{dx} \).
By recognizing the chain rule application, you transform complex operations into manageable steps, crucial for solving intricate derivatives.
The chain rule formula is \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \). This means you differentiate the outer function, evaluated at the inner function, and then multiply it by the derivative of the inner function.
In the current exercise, we apply the chain rule to the left side of our equation after taking the logarithm, \( \ln y \). It allows us to express \( \frac{d}{dx}(\ln y) \) as \( \frac{1}{y} \frac{dy}{dx} \).
By recognizing the chain rule application, you transform complex operations into manageable steps, crucial for solving intricate derivatives.
Product Rule
When two functions are multiplied, and you need to find the derivative of their product, the product rule comes into play. The product rule states that for two differentiable functions \( u(x) \) and \( v(x) \), their product's derivative \( \frac{d}{dx}[u(x)v(x)] \) is given by \( u'(x)v(x) + u(x)v'(x) \).
In our exercise, \( \ln y = -\frac{1}{x} \ln x \) involves a product on the right side. We apply the product rule here to differentiate \( -\frac{1}{x} \) and \( \ln x \) separately.
The derivative of \( -\frac{1}{x} \) using simple differentiation is \( \frac{1}{x^2} \), while the derivative of \( \ln x \) is \( \frac{1}{x} \). Using these in the product rule grants us:
In our exercise, \( \ln y = -\frac{1}{x} \ln x \) involves a product on the right side. We apply the product rule here to differentiate \( -\frac{1}{x} \) and \( \ln x \) separately.
The derivative of \( -\frac{1}{x} \) using simple differentiation is \( \frac{1}{x^2} \), while the derivative of \( \ln x \) is \( \frac{1}{x} \). Using these in the product rule grants us:
- \( \text{Derivative of the first term} \times \text{second term} \)
- \(+ \text{first term} \times \text{Derivative of the second term} \)
Natural Logarithm
The natural logarithm, written as \( \ln(x) \), is a logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. It is a widely used function in calculus due to its unique properties.
One of the pivotal features of \( \ln(x) \) is its simple derivative, \( \frac{d}{dx}(\ln x) = \frac{1}{x} \). This differentiability property makes it a favored tool when simplifying differentiation problems. In our exercise, it helps convert complex forms to linear, manageable expressions.
Using the natural logarithm transforms challenging power rules into linear calculations, as seen in the expression \( \ln(x^{-1/x}) \). This conversion can greatly simplify our task by turning multiplication into addition and bringing exponents down, facilitating straightforward differentiation.
The natural logarithm serves as a bridge to simplify and streamline the differentiation process, making seemingly complex calculations easier to solve.
One of the pivotal features of \( \ln(x) \) is its simple derivative, \( \frac{d}{dx}(\ln x) = \frac{1}{x} \). This differentiability property makes it a favored tool when simplifying differentiation problems. In our exercise, it helps convert complex forms to linear, manageable expressions.
Using the natural logarithm transforms challenging power rules into linear calculations, as seen in the expression \( \ln(x^{-1/x}) \). This conversion can greatly simplify our task by turning multiplication into addition and bringing exponents down, facilitating straightforward differentiation.
The natural logarithm serves as a bridge to simplify and streamline the differentiation process, making seemingly complex calculations easier to solve.
