Problem 25 - Calculus Single Variable

Question

Calculate the derivative with respect to \(x\) of the given expression. \(2^{\ln (x)}\)

Step-by-Step Solution

Answer

The derivative of \(2^{\ln(x)}\) is \(\frac{2^{\ln(x)} \cdot \ln(2)}{x}\).

Step 1: Rewrite the Expression

To differentiate the expression \(2^{\ln(x)}\), first, rewrite it using the property of exponents and logarithms: \(a^{b} = e^{b \cdot \ln(a)}\). Therefore, \(2^{\ln(x)}\) can be rewritten as \(e^{\ln(x) \cdot \ln(2)}\).

Step 2: Differentiate Using Chain Rule

Now, differentiate \(e^{\ln(x) \cdot \ln(2)}\) with respect to \(x\) using the chain rule. Let \(u = \ln(x) \cdot \ln(2)\). So, \(\frac{d}{dx}[e^u] = e^u \cdot \frac{du}{dx}\).

Step 3: Differentiate the Inner Function

Differentiate the inner function \(u = \ln(x) \cdot \ln(2)\) with respect to \(x\). Since \(\ln(2)\) is a constant, \(\frac{du}{dx} = \ln(2) \cdot \frac{d}{dx}[\ln(x)] = \ln(2) \cdot \frac{1}{x}\).

Step 4: Apply the Derivative to Get the Final Expression

Substitute back \(u\) and \(\frac{du}{dx}\) to find the derivative: \[ \frac{d}{dx}[e^{\ln(x) \cdot \ln(2)}] = e^{\ln(x) \cdot \ln(2)} \cdot \left(\ln(2) \cdot \frac{1}{x}\right).\]Replace \(e^{\ln(x) \cdot \ln(2)}\) with the original expression \(2^{\ln(x)}\): \[ \frac{d}{dx}[2^{\ln(x)}] = 2^{\ln(x)} \cdot \frac{\ln(2)}{x}.\]

Step 5: Simplify the Result

There are no further simplifications to be made involving the expression. Thus, the final answer is \[ \frac{d}{dx}[2^{\ln(x)}] = \frac{2^{\ln(x)} \cdot \ln(2)}{x}.\]

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. When you have a function inside another function, the chain rule lets you differentiate such expressions correctly. Suppose you have a function in the form of \( g(f(x)) \). To find the derivative \( \frac{d}{dx}[g(f(x))] \), the chain rule provides a way to do this by taking the derivative of \( g \) with respect to \( f(x) \), and then multiplying by the derivative of \( f(x) \) with respect to \( x \). In formula terms:

• The chain rule states: \( \frac{d}{dx}[g(f(x))] = g'(f(x)) \cdot f'(x) \).

In our task, the original expression \( 2^{\ln(x)} \) is rewritten as \( e^{\ln(x) \cdot \ln(2)} \). Here, \( u = \ln(x) \cdot \ln(2) \) is the inner function \( f(x) \), and \( e^u \) is the outer function \( g(u) \). We applied the chain rule by first differentiating the outer function \( e^u \) giving \( e^u \cdot \frac{du}{dx} \), and then multiplying it with the derivative of the inner function \( u \). This principle allows us to deal efficiently with complex compositions.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and have important properties that make them useful in calculus. The natural logarithmic function, denoted as \( \ln(x) \), is the logarithm with base \( e \), where \( e \approx 2.71828 \). It's vital in differentiating functions involving constants raised to variables. These properties were leveraged in the exercise.

• The basic derivative of the natural logarithm is: \( \frac{d}{dx}[\ln(x)] = \frac{1}{x} \).
• When logarithms are accompanied by constants or other expressions, they simplify the differentiation process.

In our exercise, \( \ln(x) \) forms part of an exponent in the rewrite step using \( 2^{\ln(x)} = e^{\ln(x) \cdot \ln(2)} \). The derivative of \( \ln(x) \), a vital part of solving this problem, makes it possible to apply the chain rule efficiently by recognizing the term as a separate differentiable component. This recognizes how logarithmic functions are combined and manipulated using their unique properties.

Exponential Functions

Exponential functions typically have the form \( a^x \), where \( a \) is a positive real number. The constant base \( e \) is special due to its natural properties in calculus, as it simplifies differentiation. In calculus, when dealing with differentiating expressions involving exponentials, transforming the expression can often make the calculation easier.

• For the exponential term \( e^u \), the derivative is simply \( e^u \cdot \frac{du}{dx} \) when applying the chain rule.
• This simplicity in differentiation arises from the property of the natural logarithm and its relationship with the base \( e \).

Recognizing the connection between exponentials and logarithms enables us to convert \( 2^{\ln(x)} \) to a more manageable form \( e^{\ln(x) \cdot \ln(2)} \) in our exercise. This transformation was key to leveraging the properties of exponential functions, which inherently simplifies the process of differentiation. Exponential functions are omnipresent in calculus due to their straightforward differentiation properties, helping tackle more complex functions effectively.