Problem 83 - Calculus and its applications

Question

Differentiate. $$ f(x)=\ln \left(e^{x}+1\right) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{e^x}{e^x+1} \).

Step 1: Identify the Rule for Differentiation

The function given is a composition of two functions: the natural logarithm function and an exponential function. We will use the chain rule to differentiate this composite function.

Step 2: Differentiate the Outer Function

Differentiate the outer function, the natural logarithm, with respect to its argument. If \( u = e^x + 1 \), then \( \frac{d}{du}(\ln u) = \frac{1}{u} \).

Step 3: Differentiate the Inner Function

Next, differentiate the inner function \( u = e^x + 1 \) with respect to \( x \). The derivative \( \frac{d}{dx}(e^x + 1) \) is \( e^x \).

Step 4: Apply the Chain Rule

Using the chain rule, which states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), multiply the derivatives obtained in Steps 2 and 3. This gives:\[ \frac{d}{dx}[\ln(e^x + 1)] = \frac{1}{e^x + 1} \times e^x = \frac{e^x}{e^x + 1} \]

Key Concepts

Chain Rule

When dealing with complex functions that are compositions of two or more functions, the Chain Rule is an essential tool in calculus differentiation. It is like peeling an onion, where you differentiate from the outside in.

The Chain Rule states that if you have a composite function \( f(g(x)) \), and you want to differentiate it, you do so by differentiating the outer function while keeping the inner function unchanged, and then multiplying by the derivative of the inner function itself.

• Differentiating the outer: If \( y = f(u) \) where \( u = g(x) \), then \( \frac{dy}{du} \).
• Differentiating the inner: Find \( \frac{du}{dx} \) for \( u = g(x) \).
• Apply the rule: Multiply the derivatives: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

Using the Chain Rule helps us tackle problems that are not straightforward. It is especially useful in this problem, where a logarithmic function is composed with an exponential function.

Natural Logarithm Differentiation

Natural logarithms, represented by \( \ln(x) \), have their own rules when it comes to differentiation. This makes them unique in the context of calculus differentiation.

The special nature of the natural logarithm means if you differentiate \( \ln(u) \) with respect to \( u \), you obtain \( \frac{1}{u} \). This implies that, given a natural logarithm function \( \ln(e^x + 1) \), its derivative requires differentiating the argument separately, following the Chain Rule.

• Identify the argument of the \( \ln \) function: Here, \( u = e^x + 1 \).
• Differentiate the \( \ln \) function: \( \frac{d}{du} \ln(u) = \frac{1}{u} \).
• This formula helps simplify the differentiation by reducing complex expressions into simpler ones.

Thus, the natural logarithm's differentiation is straightforward but plays a crucial role in composite functions by simplifying the outer layer of the differentiation process.

Exponential Function Differentiation

Exponential functions such as \( e^x \) have an important property: their rate of change is proportional to their current value. This leads to a unique and elegant derivative.

For an exponential function, specifically \( e^x \), the derivative remains the same, i.e., \( \frac{d}{dx}(e^x) = e^x \). This property is fundamental and provides a straightforward approach in differentiation.

• Recognize the function: Here, we have \( u = e^x + 1 \).
• Apply the derivative: The derivative of \( e^x \) is \( e^x \), and for constants like '1', it drops to zero.
• Combine this with other rules, like the Chain Rule, to tackle derivatives in composite functions.

Understanding exponential functions and their derivatives makes manipulating complex functions like \( \ln(e^x + 1) \) much more manageable. It highlights how differentiation handles both simple and complex mathematical relationships smoothly.