Problem 33 - Calculus and its applications

Question

Differentiate. $$ G(x)=\log _{9} x \cdot\left(4^{x}\right)^{6} $$

Step-by-Step Solution

Answer

The derivative is \( G'(x) = \frac{4^{6x}}{x\ln 9} + 6\log_9 x \cdot 4^{6x} \ln 4 \).

Step 1: Understanding the Expression

The function is given as \( G(x) = \log_{9}x \cdot (4^x)^6 \). This is a product of two functions: \( u(x) = \log_{9}x \) and \( v(x) = (4^x)^6 = 4^{6x} \). To differentiate this, we'll use the product rule.

Step 2: Product Rule

The product rule states that \((uv)' = u'v + uv'\). We need to find \( u' \) and \( v' \).

Step 3: Differentiating \( u(x) \)

The function \( u(x) = \log_9x \) can be rewritten using the change of base formula: \( \log_9x = \frac{\ln x}{\ln 9} \). Differentiating this gives \( u'(x) = \frac{1}{x \ln 9} \).

Step 4: Differentiating \( v(x) \)

For \( v(x) = 4^{6x} \), we can use the chain rule. Let \( k(x) = 6x \), then \( v(x) = 4^{k(x)} \). The derivative \( v'(x) = 4^{6x} \cdot \ln 4 \cdot 6 \).

Step 5: Applying Product Rule

Now substitute \( u' \,\)\( v \,\) and \( v' \,\) into the product rule formula: \( G'(x) = \left(\frac{1}{x\ln 9}\right) \cdot 4^{6x} + \log_9 x \cdot 4^{6x + 1} \cdot 6 \cdot \ln 4 \).

Step 6: Simplifying the Derivative

After applying the product rule, simplify the expression: \[ G'(x) = \frac{4^{6x}}{x\ln 9} + 6\log_9 x \cdot 4^{6x} \ln 4 \].

Key Concepts

Product Rule

When you are faced with differentiating the product of two functions, the product rule is your go-to tool. It helps us compute the derivative of a function that can be expressed as the product of two or more functions. In its basic form, if you have a function defined as the product of two functions \( u(x) \) and \( v(x) \), then the derivative, denoted as \((uv)'\), can be found using the formula:

• \((uv)' = u'v + uv'\)

This formula tells us we need to compute the derivatives of both \( u(x) \) and \( v(x) \), denoted as \( u'(x) \) and \( v'(x) \) respectively. After that, you simply plug them into the formula, where you multiply the derivative of the first function by the second function, and then add it to the first function times the derivative of the second. It's similar to product calculations in arithmetic but adjusted for functions.

In our original exercise, we identified \( u(x) = \log_9x \) and \( v(x) = 4^{6x} \). We used the product rule to find \( G'(x) \). This step ensures we correctly account for the change rates in both parts of our function.

Logarithmic Functions

Logarithmic functions are based on the concept of logarithms, which are the inverses of exponential functions. When we talk about logarithmic functions, we often see them denoted as \( \log_b(x) \), where \( b \) is the base of the logarithm. In our problem, the base is 9, so we work with \( \log_9x \).

To differentiate logarithmic functions, we might need to employ the change of base formula, which expresses \( \log_b(x) \) in terms of natural logarithms (ln). The change of base formula is:

• \( \log_b(x) = \frac{\ln x}{\ln b} \)

This allows us to use the rules of natural logarithms to differentiate. In our exercise, after converting \( \log_9(x) \) to natural logarithms, the derivative is derived using the formula for natural logs, resulting in \( \frac{1}{x \ln 9} \).

By converting to natural logarithms, it becomes straightforward to differentiate because natural logarithms have a simple derivative formula \( \ln x = \frac{1}{x} \). It's essential to use this rule correctly when dealing with logarithmic differentiation.

Exponential Functions

Exponential functions are incredibly powerful in mathematics, often represented as \( a^x \), where \( a \) is a constant and \( x \) is the variable exponent. In the context of differentiation, exponential functions have a unique feature where their rate of change is proportional to the function's current value.

The function \( 4^{6x} \) in our exercise is a prime example of exponential functions. When differentiating such functions, the chain rule is frequently utilized, especially when the exponent itself is a function of \( x \). When we have \( f(x) = a^{k(x)} \), the derivative is given by:

• \( f'(x) = a^{k(x)} \cdot \ln a \cdot k'(x) \)

Here, \( k(x) \) is the exponent function, and \( k'(x) \) is its derivative. In our problem, \( k(x) = 6x \), so \( k'(x) = 6 \). Therefore, the derivative of \( 4^{6x} \) results in \( 4^{6x} \cdot \ln 4 \cdot 6 \).

Understanding exponential differentiation is key to solving many problems in calculus, as it regularly shows up in compound interest, population growth models, and many areas of science and engineering.