Problem 345 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

\[ f'(x) = 2^x \cdot \frac{\ln(7)}{\ln(3)} \cdot [\ln(2)(x^2 - 4) + 2x] \]

Step 1: Identify the Functions and Rule to Apply

The function given is a product of two different functions: \( 2^x \) and \( \log_{3}(7^{x^2-4}) \). We need to use the product rule for differentiation, which states that if \( y = u \cdot v \), then \( y' = u'v + uv' \). Here, \( u = 2^x \) and \( v = \log_{3}(7^{x^2-4}) \).

Step 2: Differentiate \( u = 2^x \)

The derivative of an exponential function \( a^x \) with respect to \( x \) is \( a^x \ln(a) \). So, the derivative of \( u = 2^x \) is \( u' = 2^x \ln(2) \).

Step 3: Differentiate \( v = \log_{3}(7^{x^2-4}) \)

Using the change of base formula, \( \log_3(a) = \frac{\ln(a)}{\ln(3)} \). So, \( v = \frac{\ln(7^{x^2-4})}{\ln(3)} \). We can simplify \( \ln(7^{x^2-4}) = (x^2 - 4) \ln(7) \). Thus, \( v = \frac{(x^2 - 4) \ln(7)}{\ln(3)} \).

Step 4: Differentiate the Simplified \( v \)

Apply the derivative \( \frac{d}{dx} [c \, g(x)] = c \, g'(x) \) to \( v = \frac{(x^2 - 4) \ln(7)}{\ln(3)} \). The constant is \( \frac{\ln(7)}{\ln(3)} \) and \( g(x) = x^2 - 4 \), whose derivative is \( 2x \). Therefore, \( v' = \frac{\ln(7)}{\ln(3)} \cdot 2x \).

Step 5: Apply the Product Rule

Using the product rule \( y' = u'v + uv' \):- First term: \( u'v = 2^x \ln(2) \cdot \frac{(x^2 - 4) \ln(7)}{\ln(3)} \).- Second term: \( uv' = 2^x \cdot \frac{\ln(7)}{\ln(3)} \cdot 2x \).Thus, the derivative is:\[ f'(x) = 2^x \ln(2) \cdot \frac{(x^2 - 4) \ln(7)}{\ln(3)} + 2^x \cdot \frac{2x \ln(7)}{\ln(3)} \].

Step 6: Simplify the Expression

Combine the expressions for clarity:\[ f'(x) = 2^x \cdot \frac{\ln(7)}{\ln(3)} \cdot [\ln(2)(x^2 - 4) + 2x] \].

Key Concepts

Product Rule

In calculus differentiation, the product rule is essential when differentiating functions that are products of two or more sub-functions. The general rule is: if you have a function defined as the product of two functions, say \( y = u \cdot v \), the derivative of \( y \) with respect to \( x \) is given by:

• \( y' = u'v + uv' \)

To apply the product rule, you need to:

• Differently identify each part of the product as separate functions (in our problem, \( u = 2^x \) and \( v = \log_{3}(7^{x^2-4}) \)).
• Find the derivative of each function independently.
• Use the rule by multiplying each function with the derivative of the other function and summing them up.

The product rule is useful for functions that would be hard to differentiate otherwise. Understanding and applying this rule simplifies complex derivatives.

Exponential Functions

Exponential functions have the form \( a^x \), where \( a \) is a positive real number. These functions exhibit rapid growth or decay and are prevalent in many fields like population growth, radioactive decay, and finance.
For differentiation, an important result is:

• The derivative of \( a^x \) with respect to \( x \) is \( a^x \ln(a) \).

This rule arises because the base \( a \) stays constant while the exponent \( x \) varies, and natural logarithms come into play due to their properties with respect to derivatives. In our original exercise:

• We used the derivative of \( 2^x \), which is \( 2^x \ln(2) \).

Remember, exponential functions have derivatives that are products of the original function and the natural log of the base. This is a key property that distinguishes them from polynomial or rational functions.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that are composed of potentially challenging logarithms. Especially when variables are exponents, logarithmic differentiation can simplify the process.
Logarithms transform multiplicative relationships into additive ones, which are easier to handle. If the original function is \( v = \log_{3}(7^{x^2-4}) \), the steps are:

• Use the change of base formula: \( \log_3(a) = \frac{\ln(a)}{\ln(3)} \).
• Apply the property of logarithms: \( \ln(7^{x^2-4}) = (x^2 - 4) \ln(7) \), making it easier to apply standard differentiation methods.

Further simplification uses constants (here, \( \frac{\ln(7)}{\ln(3)} \)) to isolate the effect of the logarithm itself, leading to straightforward derivatives.
This method is particularly beneficial for functions with complex compositions by breaking down exponentials and products into simpler parts. By mastering logarithmic differentiation, one can handle a wide array of seemingly daunting problems with ease.