Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated derivative. \(y^{\prime}\) if \(y=e^{x / \ln x}\)

Step-by-Step Solution

Answer

The derivative is \(y^{\prime} = e^{x / \ln x} \cdot \frac{\ln x - 1}{(\ln x)^2}\).

Step 1: Understanding the Expression

The function given is in the form of an exponent with a variable in both the base and exponent: \(y = e^{x / \ln x}\). We want to find the derivative of \(y\) with respect to \(x\).

Step 2: Using the Chain Rule

Since the function is in the form \(y = e^u\), where \(u = \frac{x}{\ln x}\), we apply the chain rule. The derivative of \(e^u\) with respect to \(u\) is \(e^u\), and then multiply by the derivative of the exponent \(u\) with respect to \(x\).

Step 3: Finding the Derivative of the Exponent Function

Differentiate \(u = \frac{x}{\ln x}\). Use the quotient rule where \(u = \frac{f}{g}\), \(f = x\), and \(g = \ln x\):\[ u^{\prime} = \frac{g \cdot f^{\prime} - f \cdot g^{\prime}}{g^2} = \frac{\ln x \cdot 1 - x \cdot \frac{1}{x}}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2} \]

Step 4: Applying the Chain Rule

Substitute \(u^{\prime} = \frac{\ln x - 1}{(\ln x)^2}\) back into the chain rule for \(y = e^u\):\[ y^{\prime} = e^{x / \ln x} \cdot \frac{\ln x - 1}{(\ln x)^2} \]

Step 5: Concluding the Derivative

The expression \(y = e^{x / \ln x}\) differentiates to:\[ y^{\prime} = e^{x / \ln x} \cdot \frac{\ln x - 1}{(\ln x)^2} \]

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of composite functions, or functions within functions. When you have a function that is composed as one inside another, like in our exercise with the form \(e^u\), the chain rule becomes invaluable.

In the given problem, we recognize that \(y = e^{x / \ln x}\) is a composite function. The inner function \(u\) is \(\frac{x}{\ln x}\), and \(e^u\) is the outer function. According to the chain rule, to find the derivative of a composition, you take the derivative of the outer function with respect to the inner function and multiply it by the derivative of the inner function with respect to the variable, which in this case is \(x\).

With this strategy:

• The derivative of \(e^u\) with respect to \(u\) is still \(e^u\), as the derivative of an exponential function with base \(e\) is itself.
• We then multiply by the derivative of \(u = \frac{x}{\ln x}\), which we will find using the quotient rule.

Mastering the Quotient Rule

The quotient rule is applied when you need to find the derivative of a function that is the division of two other functions. In this exercise, we have \(u = \frac{x}{\ln x}\), and we need to apply the quotient rule to find its derivative.

The quotient rule formula is:

• Given \(u = \frac{f}{g}\), the derivative \(u'\) is \(\frac{g \cdot f' - f \cdot g'}{g^2}\).

In our function:

• \(f = x\) with \(f' = 1\), since the derivative of \(x\) with respect to \(x\) is 1.
• \(g = \ln x\) with \(g' = \frac{1}{x}\), as the derivative of \(\ln x\) is \(\frac{1}{x}\).

Plugging these into the quotient rule formula, we compute:

• \(u^{\prime} = \frac{(\ln x) \cdot 1 - x \cdot \frac{1}{x}}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2}\).

Through careful application of the quotient rule, we now have the derivative of the exponent function \(u\). This crucial step allows us to fully apply the chain rule to find \(y'\).

Exploring Exponential Functions

Exponential functions are where the variable appears in the exponent, typically in the form \(a^x\) or \(e^x\). These functions have unique properties that make their differentiation very straightforward.

In the function \(y = e^{x / \ln x}\), \(e\) is the base of our natural exponential function. An essential property of the natural exponential function \(e^x\) is that its derivative is itself, \(d/dx(e^u) = e^u\). This simplicity arises from the constant rate of growth of the exponential function.

When differentiating a natural exponential function with respect to a different function, such as \(e^u\), we use the chain rule to account for the inner structure of \(u\):

• First, calculate the derivative of the outer function, which is \(e^u\) for our problem.
• Then, multiply it by the derivative of the exponent function \(u\) with respect to \(x\), which we've already found using the quotient rule.

In conclusion, for \(y = e^{x / \ln x}\), the final form of the derivative is \(y^{\prime} = e^{x / \ln x} \cdot \frac{\ln x - 1}{(\ln x)^2}\), demonstrating the power and elegance of exponential differentiation when combined with the chain and quotient rules.