Problem 62 - Calculus and its applications

Question

Use the Chain Rule, implicit differentiation, and other techniques to differentiate each function given. $$ f(x)=x^{e^{x}}, \text { for } x>0 $$

Step-by-Step Solution

Answer

The derivative is \(f'(x) = e^x x^{e^x-1} + \ln(x) e^x x^{e^x}\).

Step 1: Understand the Structure of the Function

The given function is in the form of a power function where the base and the exponent both depend on the variable \(x\). Specifically, the function is \(f(x) = x^{e^x}\). Differentiating such a function requires the use of logarithmic differentiation.

Step 2: Apply Logarithmic Differentiation

Begin by taking the natural logarithm of both sides of the equation: \[ \ln(f(x)) = \ln(x^{e^x}).\]Using the property of logarithms \(\ln(a^b) = b \ln(a)\), rewrite the equation as:\[ \ln(f(x)) = e^x \ln(x).\]

Step 3: Differentiate Both Sides Implicitly

Differentiate both sides with respect to \(x\):\[ \frac{d}{dx}[\ln(f(x))] = \frac{d}{dx}[e^x \ln(x)].\]The left side becomes \(\frac{1}{f(x)} \cdot f'(x)\) by the chain rule since \(f(x)\) is a function of \(x\). Apply the product rule to the right side:\[ e^x \cdot \frac{d}{dx}[\ln(x)] + \ln(x) \cdot \frac{d}{dx}[e^x]\]This simplifies to:\[ e^x \cdot \frac{1}{x} + \ln(x) \cdot e^x\]Thus we have: \[ \frac{1}{f(x)} \cdot f'(x) = \frac{e^x}{x} + \ln(x) e^x.\]

Step 4: Solve for \(f'(x)\)

Multiply both sides by \(f(x)\) to solve for \(f'(x)\):\[ f'(x) = f(x) \left(\frac{e^x}{x} + \ln(x) e^x\right).\]Recall that \(f(x) = x^{e^x}\), so:\[ f'(x) = x^{e^x} \left(\frac{e^x}{x} + \ln(x) e^x\right).\]Further simplify:\[ f'(x) = e^x x^{e^x - 1} + \ln(x) e^x x^{e^x}.\]This is the derivative of the function \(f(x) = x^{e^x}\).

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus used for differentiating compositions of functions. When you have a function within another function, such as \((g(x) = f(h(x)))\), the Chain Rule allows you to find the derivative using the formula:

• \(g'(x) = f'(h(x)) \cdot h'(x)\)

This rule is extremely powerful when dealing with complex expressions where a direct differentiation isn't feasible.
In the context of our exercise, the Chain Rule comes into play when differentiating the logarithm of the function. Consider the expression \(\ln(f(x)) = e^x \ln(x)\). Here, you'll notice that \(f(x)\) is itself a complex wrapped up expression \(x^{e^x}\).
To differentiate \(\ln(f(x))\), the Chain Rule helps internalize the derivative because \(f(x)\) is a function of \(x\).
By utilizing the Chain Rule, you unravel the exponents and other nested functions step-by-step, eventually simplifying the derivative process for complex power functions. This concept is crucial to successfully manage derivative calculations where simple methods don't suffice.

Implicit Differentiation

Implicit Differentiation arises when you have equations not explicitly solved for one variable in terms of another, meaning variables are intertwined, such as in equations of circles or curves.
To differentiate implicitly, you take the derivative of both sides of the equation with respect to a variable, often \(x\).

• Derive each term by applying standard differentiation rules, treating other variables as implicit functions of \(x\).
• Remember to multiply by the derivative of these implicit functions, which emerges notably through the Chain Rule.

In our exercise, after applying logarithmic differentiation, we use implicit differentiation to handle the derivative of the logarithmic expression: \[ e^x \ln(x)\].
Here, the implicit nature stems from the entwined \(e^x\) and \(\ln(x)\), necessitating distinct attention when deriving.
Differentiating implicitly allows us to manage complex interactivities within equations, leading smoothly to deriving relationships even when functions are not neatly isolated by variable.

Logarithmic Differentiation

Logarithmic Differentiation is a technique used when dealing with functions raised to complicated exponents, usually involving variable bases and exponents. It transforms multiplicative relationships into additive ones by leveraging logarithmic properties.
Here's how it works:

• Take the natural logarithm of both sides of your function equation.
• Use the logarithm property \( \ln(a^b) = b \ln(a) \) to simplify.
• Differentiate both sides using implicit methods, applying derivatives to logarithmic expressions.

For our exercise \( f(x) = x^{e^x} \), logarithmic differentiation conveniently deals with the power function requiring derivation. By converting the expression to \( \ln(f(x)) = e^x \ln(x) \), you ease the complex differentiation task into simpler, manageable steps.
Applying derivatives to each component separately then provides a path to finding \( f'(x) \).
Logarithmic differentiation simplifies differentiation tasks that at first glance, appear daunting, enabling clean resolutions for functions cocooned within powers.