Chapter 1: Problem 36
For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$ y=e^{-\ln x} $$
Short Answer
Expert verified
The derivative is \( \frac{dy}{dx} = -\frac{1}{x^2} \).
Step by step solution
01
Understand the Given Function
The given function to differentiate is \( y = e^{-\ln x} \). This is an exponential function where the exponent is \(-\ln x\).
02
Simplify the Exponent
Recognize that the expression \( e^{-\ln x} \) can be further simplified using properties of logarithms and exponentials. We know that \( e^{\ln a} = a \). Thus, \( e^{-\ln x} = e^{\ln x^{-1}} = x^{-1} = \frac{1}{x} \). So, the function simplifies to \( y = \frac{1}{x} \).
03
Take the Natural Logarithm on Both Sides
To use the differentiation technique involving \( \ln y \), start by taking the natural logarithm of both sides: \( \ln y = \ln \left( \frac{1}{x} \right) \).
04
Differentiate "Log y" with Respect to x
Differentiate both sides of the equation \( \ln y = -\ln x \) with respect to \( x \). Recall the differentiation rule for \( \ln u \): \( \frac{d}{dx} \ln u = \frac{u'}{u} \). This becomes \( \frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx} \) and \( \frac{d}{dx}(-\ln x) = -\frac{1}{x} \).
05
Solve for dy/dx
Set the derivatives equal to each other: \( \frac{1}{y} \cdot \frac{dy}{dx} = -\frac{1}{x} \). Rearrange to solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = -\frac{y}{x} \). Since \( y = \frac{1}{x} \), substitute to find \( \frac{dy}{dx} = -\frac{1/x}{x} = -\frac{1}{x^2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Functions
Exponential functions are a special type of mathematical function where the variable appears in the exponent. A typical exponential function can be expressed as \( y = a^x \), where \( a \) is a constant base, and \( x \) is the exponent. In our exercise, we dealt with the function \( y = e^{-\ln x} \), which looked more complex due to the composition involving a natural logarithm.
One of the key traits of an exponential function is its rapid growth or decay, depending on whether the base is greater than or less than one. The constant \( e \) (approximately 2.718) is often used as the base when dealing with natural logarithms, which simplifies many mathematical expressions due to its unique properties.
In the given exercise, the function simplifies from \( e^{-\ln x} \) to \( \frac{1}{x} \) by recognizing that \( e^{\ln x^{-1}} = x^{-1} \). This step highlights the importance of understanding the properties of exponential and logarithmic functions to simplify expressions correctly.
One of the key traits of an exponential function is its rapid growth or decay, depending on whether the base is greater than or less than one. The constant \( e \) (approximately 2.718) is often used as the base when dealing with natural logarithms, which simplifies many mathematical expressions due to its unique properties.
In the given exercise, the function simplifies from \( e^{-\ln x} \) to \( \frac{1}{x} \) by recognizing that \( e^{\ln x^{-1}} = x^{-1} \). This step highlights the importance of understanding the properties of exponential and logarithmic functions to simplify expressions correctly.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a logarithm with the base \( e \). It is widely used in calculus because it simplifies many derivatives and integrals. In the context of solving differential equations, natural logarithms often help in transforming complex multiplication and exponentiation into simpler addition and subtraction.
In our original problem, we leverage the natural logarithm of both sides of the equation \( y = e^{-\ln x} \) to make the differentiation process more straightforward. By applying \( \ln \), the function became \( \ln y = -\ln x \). This transformation uses the logarithmic property \( \ln \left( \frac{a}{b} \right) = \ln a - \ln b \), which helps simplify the equation for differentiation.
Understanding how to manipulate functions using logarithms can greatly reduce the complexity involved in differentiation, as seen in this exercise.
In our original problem, we leverage the natural logarithm of both sides of the equation \( y = e^{-\ln x} \) to make the differentiation process more straightforward. By applying \( \ln \), the function became \( \ln y = -\ln x \). This transformation uses the logarithmic property \( \ln \left( \frac{a}{b} \right) = \ln a - \ln b \), which helps simplify the equation for differentiation.
Understanding how to manipulate functions using logarithms can greatly reduce the complexity involved in differentiation, as seen in this exercise.
Derivative Rules
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. When working with exponential and logarithmic functions, there are specific derivative rules that are invaluable.
For a natural logarithm \( \ln u \), the derivative is \( \frac{d}{dx}(\ln u) = \frac{u'}{u} \), meaning you derive "\( u \)" with respect to \( x \) and then divide by \( u \). In our exercise, this rule allowed us to differentiate \( \ln y \) efficiently: \( \frac{1}{y} \frac{dy}{dx} = -\frac{1}{x} \).
Once this equation was derived, solving for \( \frac{dy}{dx} \) involved algebraic manipulation resulting in \( \frac{dy}{dx} = -\frac{1}{x^2} \). It is crucial to be comfortable with these derivative rules to tackle the differentiation of complex functions efficiently, especially when they involve compositions like \( e^{\ln} \) and other logarithmic expressions.
For a natural logarithm \( \ln u \), the derivative is \( \frac{d}{dx}(\ln u) = \frac{u'}{u} \), meaning you derive "\( u \)" with respect to \( x \) and then divide by \( u \). In our exercise, this rule allowed us to differentiate \( \ln y \) efficiently: \( \frac{1}{y} \frac{dy}{dx} = -\frac{1}{x} \).
Once this equation was derived, solving for \( \frac{dy}{dx} \) involved algebraic manipulation resulting in \( \frac{dy}{dx} = -\frac{1}{x^2} \). It is crucial to be comfortable with these derivative rules to tackle the differentiation of complex functions efficiently, especially when they involve compositions like \( e^{\ln} \) and other logarithmic expressions.
