Problem 15 - Single Variable Calculus

Question

Find the differential of the function at the indicated number. $$ f(x)=e^{x}+\ln (1+x) ; \quad x=0 $$

Step-by-Step Solution

Answer

The differential of the function \(f(x) = e^x + \ln (1+x)\) at the point \(x=0\) is 2.

Step 1: Find the derivative of the function f(x)

Recall the derivative rules of the exponential function and the natural logarithm function: 1. If \(y = e^x\), then \(y' = e^x\). 2. If \(y = \ln(u)\), then \(y' = \frac{1}{u} \cdot u'\), where u is a function of x. Now, let's use these rules to find the derivative of \(f(x) = e^x + \ln (1+x)\). The first term is \(e^x\), so its derivative is also \(e^x\). In the second term, we have \(\ln (1+x)\). Here, u = 1+x and \(u' = 1\). So, applying the second rule, the derivative of \(\ln(1+x)\) is \(\frac{1}{1+x} \cdot 1 =\frac{1}{1+x}\). We can now write the derivative of the whole function: \(f'(x) = e^x + \frac{1}{1+x}\)

Step 2: Evaluate the derivative at x=0

Now that we have the derivative of the function, we need to evaluate it at the given value \(x=0\). This will give us the differential of the function at this point. \(f'(0) = e^0 + \frac{1}{1+0}\) Recall that \(e^0=1\), so our expression becomes: \(f'(0) = 1 + \frac{1}{1+0}=1+\frac{1}{1}=1+1\)

Step 3: Calculate the value of the differential

Finally, the sum of the expression above results in: \(f'(0) = 2\) Thus, the differential of the function \(f(x) = e^x + \ln (1 + x)\) at the point \(x=0\) is 2.

Key Concepts

Exponential Function

The exponential function is a fundamental concept in mathematics, represented by the expression \(e^x\), where \(e\) is a constant approximately equal to 2.71828. It is the base of the natural logarithm. One of the key properties of the exponential function is that its growth rate is proportional to its current value, making it incredibly useful across different fields such as finance, biology, and physics.
The derivative of an exponential function, \(e^x\), is unique because it is the only function that is its own derivative. This means:

• The rate of change (or slope) of \(e^x\) at any point is \(e^x\) itself.
• This property makes it straightforward to work with in differential calculus, simplifying the process of finding derivatives.

Overall, understanding the exponential function and its derivative property is crucial in solving more complex calculus problems.

Natural Logarithm

The natural logarithm, noted as \(\ln(x)\), is the inverse operation of raising \(e\) to a power. It answers the question: **To what power must we raise \(e\) to obtain \(x\)?**

Natural logarithms are essential in converting exponential growth into linear growth, which is easier to analyze and interpret. The most significant derivative-related property for natural logarithms is:

• The derivative of \(\ln(u)\) is \(\frac{1}{u} \times u'\), where \(u\) is a function of \(x\). This rule is essential for finding the derivatives of compositions of logarithmic functions.

As with the exponential function, understanding natural logarithms is crucial when dealing with equations involving \(e\) due to their inverse relationship. In our example problem, using this derivative rule for \(\ln(1+x)\) greatly simplifies the calculations.

Derivative Rules

Derivative rules are a set of guidelines that help us find the derivative of a function efficiently. Derivatives measure how a function changes as its input changes.
Some key derivative rules that are often used include:

• Power Rule: For \(f(x) = x^n\), the derivative \(f'(x) = nx^{n-1}\).
• Product Rule: For \(f(x) = u(x)v(x)\), the derivative is \(f'(x) = u'(x)v(x) + u(x)v'(x)\).
• Quotient Rule: For \(f(x) = \frac{u(x)}{v(x)}\), the derivative is \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}\).

For the problem at hand, we primarily use the rules for the derivatives of the exponential function and the natural logarithm. By mastering these rules, understanding and finding derivatives becomes a much simpler task.

Chain Rule

The chain rule is a fundamental derivative principle that deals with the derivative of composite functions. When a function is made up by combining other functions, the chain rule makes it possible to differentiate the composite function without having to simplify it.

The chain rule can be succinctly stated as:

• If you have a composite function \(f(g(x))\), then its derivative is \((f \circ g)'(x) = f'(g(x)) \cdot g'(x)\).

In simpler terms, the derivative of a composite function is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function.

The chain rule is highly applicable in scenarios involving exponential functions and natural logarithms, especially when they involve variable expressions.
In our problem, the application of the chain rule is crucial for accurately finding the derivative of \(\ln(1+x)\), where we determine \(u\) as \(1+x\), then find and multiply its derivative \(u'\), ensuring all components of the composite function are accounted for.