Question

Suppose that \(f(x)>0\) for all \(x\) and that \(f\) and \(g\) are differentiable. Use the identity \(f^{g}=e^{g \ln f}\) and the chain rule to find the derivative of \(f^{g}\).

Short answer

The derivative is \( f^g \cdot \left(g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)}\right) \).

Step-by-step solution

Express the Function Using Exponentiation

We begin by rewriting the function \( f^g \) using the given identity: \( f^g = e^{g \ln f} \). This allows us to work with the exponential form, which is easier to differentiate.

Differentiate the Exponential Function

To differentiate \( e^{g(x) \ln f(x)} \), we apply the chain rule. According to the chain rule, the derivative of \( e^u \) with respect to \( x \) is \( e^u \cdot \frac{du}{dx} \), where \( u = g(x) \ln f(x) \).

Find the Derivative of the Inner Function

For the inner function \( u = g(x) \ln f(x) \), we use the product rule to differentiate: \( \frac{du}{dx} = g'(x) \ln f(x) + g(x) \cdot \frac{d}{dx}(\ln f(x)) \).

Differentiate \(\ln f(x)\)

Differentiating \( \ln f(x) \) with respect to \( x \) gives \( \frac{f'(x)}{f(x)} \). This follows from the derivative rule for natural logarithms.

Substitute Back into the Derivative Expression

Substitute \( \frac{du}{dx} = g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)} \) back into the derivative of the exponential function to get: \( \frac{d}{dx}(e^{g(x) \ln f(x)}) = e^{g(x) \ln f(x)} \cdot \left(g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)}\right) \).

Solve for the Derivative of \( f^g \)

Remember \( f^g = e^{g \ln f} \). Therefore, the derivative \( \frac{d}{dx}(f^g) \) is \( f^g \cdot \left(g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)}\right) \). This is the final expression for the derivative of \( f^g \).

Key concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus. It helps in finding the derivative of composite functions, which are functions inside another function. For example, if you have a function expressed as \( h(x) = f(g(x)) \), the chain rule tells us:

• The derivative of \( h(x) \) is \( f'(g(x)) \cdot g'(x) \).

This means you first differentiate the outer function while keeping the inner function unchanged, then you multiply by the derivative of the inner function. In the context of our problem with \( e^{g(x) \ln f(x)} \), the entire expression \( u = g(x) \ln f(x) \) is considered the inner function. Therefore, by applying the chain rule, we differentiate \( e^u \) as:

• \( e^u \cdot \frac{du}{dx} \)

Product Rule

The Product Rule is essential when dealing with products of two functions. It states that if a function \( u(x) \) is composed of the product of two other functions, say \( g(x) \) and \( \ln f(x) \), then the derivative is given by:

• \( \frac{d}{dx}(g(x) \ln f(x)) = g'(x) \ln f(x) + g(x) \cdot \frac{d}{dx}(\ln f(x)) \)

This rule is especially helpful for our problem, where we need to differentiate \( g(x) \ln f(x) \). We apply the product rule to handle each part of the expression separately, ensuring that no part goes undifferentiated:

• First, differentiate \( g(x) \) to get \( g'(x) \ln f(x) \).
• Secondly, differentiate \( \ln f(x) \), which typically results in \( \frac{f'(x)}{f(x)} \).

Exponential Form

The exponential form \( e^{g \ln f} \) is a useful transformation. It simplifies the differentiation process when dealing with expressions like \( f^g \). This technique utilizes properties of exponentials to express complex functions in a form that is easier to work with.

• The main idea is to rewrite \( f^g \) as \( e^{g \ln f} \) because the derivative of an exponential function is straightforward when combined with the chain rule.

By expressing the term \( f^g \) in terms of \( e \), it becomes easier to apply rules of differentiation to find derivatives efficiently. The base \( e \) has the property that its derivative remains unchanged, multiplying only by the derivative of the exponent.

Natural Logarithm

The natural logarithm \( \ln f(x) \) is a common concept in calculus. It's the inverse of the exponential function and has distinct properties that make it attractive for differentiation.

• When you differentiate \( \ln f(x) \), you use the formula: \( \frac{1}{f(x)} \cdot f'(x) = \frac{f'(x)}{f(x)} \).

This rule gives us a straightforward way to differentiate logarithmic expressions. Knowing this, we can easily find how clips of functions like \( \ln f(x) \) contribute to the derivative of more complex expressions like \( f^g \).

• It simplifies the product rule by providing a neat way to handle the logarithmic part of the function.

Understanding how natural logarithms work is crucial for dealing with exponential expressions, especially when using them to simplify differentiation tasks.