Problem 19 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find \(f^{\prime}(x) .\) Do these problems without using the Quotient Rule. $$ f(x)=\ln \left(e^{x}+x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=\ln \left(e^{x}+x^{2}\right)\) is \(f^{\prime}(x) = \frac{e^{x}+ 2x}{e^{x}+x^{2}}\).

Step 1: Identify the outer and inner functions

The outer function is \(\ln(u)\) where \(u = e^{x}+x^{2}\) and inner function is \(e^{x}+x^{2}\).

Step 2: Find the derivative of the outer function

The derivative of the outer function \(f(u) = \ln(u)\), by the rule of natural logarithm differentiation, is \(f'(u) = 1/u = 1/(e^{x}+x^{2})\).

Step 3: Find the derivative of the inner function

The derivative of the inner function \(u = e^{x}+x^{2}\) is \(u' = e^{x} + 2x\), by using the power rule for \(x^{2}\) and the rule for the derivative of the exponential function.

Step 4: Apply the Chain Rule

By the chain rule, the derivative of the composition of these functions is \(f'(x) = f'(u)u' = (1/(e^{x}+x^{2}))*(e^{x}+2x)\).

Key Concepts

Chain Rule in Calculus

The Chain Rule is a fundamental technique in calculus that allows us to differentiate composite functions. A composite function is simply a function nested within another function. The Chain Rule states that if you have a function \( y = f(g(x)) \), then the derivative \( y' \) with respect to \( x \) is found by multiplying the derivative of the outer function evaluated at the inner function, \( f'(g(x)) \), by the derivative of the inner function, \( g'(x) \). This can be formulated as:

• \( y' = f'(g(x)) \cdot g'(x) \)

In simpler terms, think of peeling layers of an onion, where you differentiate the outer layer and multiply it by the differentiated inner part. This rule is crucial for handling functions like \(f(x)=\ln(e^{x}+x^{2})\), where we have the natural logarithm function as the outer function and \(e^{x}+x^{2}\) as the inner function. Using the Chain Rule, we can efficiently find the derivative of such compositions.

Natural Logarithm Differentiation

Differentiating natural logarithms is a technique commonly encountered in calculus. The natural logarithm function is denoted as \( \ln(x) \), and its derivative is straightforward:

• The derivative of \( \ln(x) \) is \( 1/x \).

This comes in handy because if the argument of the logarithm is a bit more complex, we apply this rule alongside the Chain Rule. For example, if we have \( f(x) = \ln(u) \), and \( u \) is a function of \( x \) itself, then the derivative is expressed as:

• \( f'(x) = \frac{1}{u} \cdot u' \)

In our exercise, using this concept, we identified the outer function as \( \ln(u) \) with \( u = e^{x} + x^{2} \), and found its derivative \( f'(u) = \frac{1}{e^{x} + x^{2}} \).

Exponential Function Derivative

The exponential function \( e^{x} \) is one of the most frequently used functions in calculus, owing to its special property: it is its own derivative. This means the derivative of \( e^{x} \) is simply \( e^{x} \).
This property makes exponential functions extremely easy to differentiate. In composite functions, like our example \( f(x) = \ln(e^{x} + x^{2}) \), the derivative of the exponential function is important when applying the Chain Rule.
Within the expression \( e^{x} + x^{2} \), we find \( e^{x} \)'s derivative to be \( e^{x} \) and combined this with the derivative of \( x^{2} \) to find \( u' \).This method highlights the elegant simplicity of exponential differentiation.

Power Rule in Calculus

The Power Rule is a key rule for differentiating functions of the form \( x^n \), where \( n \) is a constant. The Power Rule states:

• The derivative of \( x^n \) is \( nx^{n-1} \).

This means if you have a term like \( x^2 \), its derivative would be \( 2x \) according to this rule. The Power Rule is especially handy for polynomials and often works in tandem with other differentiation rules like the Chain Rule.
In our problem, when differentiating the expression \( e^{x} + x^{2} \), the term \( x^{2} \) was simplified using the Power Rule to \( 2x \). By understanding the Power Rule, we can seamlessly tackle polynomial terms within larger expressions.