Question

Find \(f^{\prime}(x)\) for each function. $$ f(x)=2^{x} \cdot \log _{3} 7^{x^{2}-4} $$

Short answer

\(f'(x) = 2^x \log_3(7)[\ln(2) (x^2-4) + 2x]\).

Step-by-step solution

Identify the Functions

The given function is a product of two functions: 1. The exponential function, \( g(x) = 2^x \).2. The logarithmic function, \( h(x) = \log_3(7^{x^2 - 4}) \). We'll need to apply the product rule to find \( f'(x) = (g \cdot h)'(x) \).

Apply the Product Rule

According to the product rule, for two functions \(g(x)\) and \(h(x)\), the derivative \((g \cdot h)'(x)\) is given by:\[(g \cdot h)'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x) \]We need to find both \(g'(x)\) and \(h'(x)\).

Differentiate the Exponential Function

The derivative of \(g(x) = 2^x\) is given by:\[g'(x) = \frac{d}{dx}[2^x] = 2^x \ln(2).\]

Simplify the Logarithmic Function

Use the properties of logarithms to simplify \(h(x)\):\[ h(x) = \log_3(7^{x^2-4}) = (x^2-4) \cdot \log_3(7).\]

Differentiate the Logarithmic Function

Now, find the derivative of the simplified \(h(x) = (x^2-4) \cdot \log_3(7) \):\[h'(x) = \frac{d}{dx}[(x^2-4) \log_3(7)] = 2x \cdot \log_3(7). \]

Substitute into the Product Rule

Substitute \(g(x)\), \(g'(x)\), \(h(x)\), and \(h'(x)\) back into the product rule:\[ f'(x) = 2^x \ln(2) \cdot \log_3(7^{x^2-4}) + 2^x \cdot 2x \cdot \log_3(7).\]

Simplify the Result

Combine the terms and simplify:\[f'(x)= 2^x \log_3(7)[\ln(2) (x^2-4) + 2x].\]

Key concepts

Product Rule in Calculus

The product rule is a fundamental principle used in calculus to find the derivative of a function that is the product of two other functions. This rule is particularly handy because it applies to any situation where two functions are multiplied together.
The product rule states that if you have a function that can be expressed as the product of two functions, such as \( f(x) = g(x) \cdot h(x) \), then the derivative of \( f(x) \), denoted as \( f'(x) \), is given by:

• \( f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x) \)

Essentially, this means you first take the derivative of the first function and multiply it by the second function. Then, take the derivative of the second function and multiply it by the first function. Finally, add these two products together to obtain the derivative of the entire product.
This approach simplifies the process of differentiation when dealing with products of functions, as seen in our example with an exponential and a logarithmic function.

Exponential Functions Differentiation

Differentiating exponential functions involves a straightforward yet significant rule. Exponential functions have the general form \( a^x \), where \( a \) is a constant. The derivative of such an exponential function, \( g(x) = a^x \), is determined by the formula:

• \( g'(x) = a^x \ln(a) \)

This formula arises from the unique property of exponential functions, where their rate of change is proportional to their current value. In our example with \( g(x) = 2^x \), applying this rule gives us the derivative \( g'(x) = 2^x \ln(2) \).
The natural logarithm, \( \ln(a) \), arises when differentiating because it reflects the continuous growth rate of the exponential function expressed on the natural base \( e \). Understanding this concept is crucial for accurately solving problems that involve exponential functions in calculus.

Logarithmic Functions Differentiation

Logarithmic functions are the inverse of exponential functions, and their differentiation involves a distinct yet related set of rules. A function of the form \( \log_b(x) \) can be differentiated using the rule:

• \( \frac{d}{dx} [\log_b(u)] = \frac{1}{u \ln(b)} \cdot u'(x) \)

In cases where the argument of the logarithm is a power or a composite function, properties of logarithms can be used to simplify the expression before differentiation. For instance, the logarithm of a power, \( \log_b(u^v) = v \cdot \log_b(u) \), is simplified accordingly.
In the original exercise, the logarithmic function \( h(x) = \log_3(7^{x^2-4}) \) was simplified using: \( h(x) = (x^2-4) \cdot \log_3(7) \). The derivative then becomes \( h'(x) = 2x \cdot \log_3(7) \), utilizing the constant multiple and power rules.
Mastering these differentiation techniques enables one to handle various calculus problems effortlessly, involving complex expressions with logarithms.