Chapter 1: Problem 31
For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$ y=e^{(e x)} $$
Short Answer
Expert verified
\( \frac{dy}{dx} = e^{(e^x + x)} \).
Step by step solution
01
Natural Logarithm of Both Sides
Start by taking the natural logarithm of both sides of the equation to simplify differentiation. So, we have:\[\ln y = \ln(e^{(e^x)})\] Applying the property of logarithms, \( \ln(a^b) = b \ln a \), we get:\[\ln y = e^x \cdot \ln e \]Since \( \ln e = 1 \), it simplifies further to:\[\ln y = e^x\]
02
Differentiate Both Sides
Differentiate both sides of the equation with respect to \(x\). Remember that \(y\) is a function of \(x\), so you will use implicit differentiation.Differentiating the left side using the chain rule, we have:\[\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}\]Differentiating the right side, we get:\[\frac{d}{dx}(e^x) = e^x\] So the equation becomes:\[\frac{1}{y} \frac{dy}{dx} = e^x\]
03
Solve for \( \frac{dy}{dx} \)
To find \( \frac{dy}{dx} \), multiply both sides of the equation by \(y\):\[\frac{dy}{dx} = y \cdot e^x\]Recall that \( y = e^{(e^x)} \). Substitute \( y \) back in:\[\frac{dy}{dx} = e^{(e^x)} \cdot e^x\]
04
Simplify the Derivative
Combine the exponents using the property \( e^a \cdot e^b = e^{a+b} \): \[\frac{dy}{dx} = e^{(e^x + x)}\] This is the final expression for the derivative \( \frac{dy}{dx} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Implicit Differentiation
Implicit differentiation is a powerful technique used when dealing with equations that are not solved explicitly for one variable in terms of another but are still linked. Unlike explicit functions where you clearly have a form like \( y = f(x) \), implicit relations intertwine variables, like those expressed as \( F(x, y) = 0 \).
In the solution, we take the derivative of \( \ln y = e^x \) using implicit differentiation. Remember that here, \( y \) is not expressed directly as a simple function of \( x \). Thus, when differentiating \( \ln y \), you regard \( y \) as a function of \( x \), and apply the chain rule, giving \( \frac{1}{y} \cdot \frac{dy}{dx} \).
This step is crucial because it allows us to differentiate non-explicit relationships by acknowledging derivatives also depend on other derivatives within the equation. It's especially useful in situations where assigning an explicit form is impractical.
In the solution, we take the derivative of \( \ln y = e^x \) using implicit differentiation. Remember that here, \( y \) is not expressed directly as a simple function of \( x \). Thus, when differentiating \( \ln y \), you regard \( y \) as a function of \( x \), and apply the chain rule, giving \( \frac{1}{y} \cdot \frac{dy}{dx} \).
This step is crucial because it allows us to differentiate non-explicit relationships by acknowledging derivatives also depend on other derivatives within the equation. It's especially useful in situations where assigning an explicit form is impractical.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base of Euler's number \( e \), which is approximately 2.718. It simplifies many expressions in calculus, particularly those involving powers and exponential functions, due to its property \( \ln(e) = 1 \).
In this exercise, we utilize the property \( \ln(a^b) = b \ln(a) \) to simplify the expression \( \ln(e^{e^x}) \) to \( e^x \cdot \ln(e) \). This simplifies further to \( e^x \) because \( \ln(e) = 1 \).
This logarithmic property helps simplify differentiation challenges by reducing complex products and powers into more manageable linear combinations. It underscores the importance of understanding fundamental logarithmic identities in calculus.
In this exercise, we utilize the property \( \ln(a^b) = b \ln(a) \) to simplify the expression \( \ln(e^{e^x}) \) to \( e^x \cdot \ln(e) \). This simplifies further to \( e^x \) because \( \ln(e) = 1 \).
This logarithmic property helps simplify differentiation challenges by reducing complex products and powers into more manageable linear combinations. It underscores the importance of understanding fundamental logarithmic identities in calculus.
Chain Rule
The chain rule is a quintessential tool in calculus used to differentiate compositions of functions. When you have a function nested inside another, the chain rule helps in finding the derivative by differentiating the outer and inner functions in sequence. The formula is often presented as \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).
In the solution, the derivative of \( \ln y \) involves using the chain rule. As \( y \) is itself a function of \( x \), we apply the chain rule as: \( \frac{d}{dx}(\ln(y)) = \frac{1}{y} \cdot \frac{dy}{dx} \). This approach allows us to systematically account for the inner function \( y \) itself depending on \( x \).
By mastering the chain rule, you gain the ability to unravel even the most complex functional compositions, an essential skill across mathematics and applied sciences.
In the solution, the derivative of \( \ln y \) involves using the chain rule. As \( y \) is itself a function of \( x \), we apply the chain rule as: \( \frac{d}{dx}(\ln(y)) = \frac{1}{y} \cdot \frac{dy}{dx} \). This approach allows us to systematically account for the inner function \( y \) itself depending on \( x \).
By mastering the chain rule, you gain the ability to unravel even the most complex functional compositions, an essential skill across mathematics and applied sciences.
Exponential Function
The exponential function is among the most significant in mathematics, defined by \( e^x \) where \( e \) is Euler's number, approximately 2.718. It's unique due to its self-replicating derivative property: differentiating \( e^x \) yields \( e^x \).
In our exercise, the function \( y = e^{(e^x)} \) is differentiated by first simplifying through logarithms and then derivative determination using implicit methods. When found that \( y \) equals to \( e^{(e^x)} \), it complicates polynomial-like growth in differentiation unless applying specialized techniques such as logs and implicit differentiation.
Understanding the traits and behaviors of exponential functions is crucial, especially in natural phenomena modeling where exponential increases and compound interest models are routine. Exponential functions represent exponential growth or decay processes, key concepts in both pure and applied disciplines.
In our exercise, the function \( y = e^{(e^x)} \) is differentiated by first simplifying through logarithms and then derivative determination using implicit methods. When found that \( y \) equals to \( e^{(e^x)} \), it complicates polynomial-like growth in differentiation unless applying specialized techniques such as logs and implicit differentiation.
Understanding the traits and behaviors of exponential functions is crucial, especially in natural phenomena modeling where exponential increases and compound interest models are routine. Exponential functions represent exponential growth or decay processes, key concepts in both pure and applied disciplines.
