Problem 59 - Single Variable Calculus

Question

Find the derivative of the function. $$ h(x)=\left(2^{x}+3^{-x}\right)^{6} $$

Step-by-Step Solution

Answer

The derivative of the function \(h(x) = (2^x + 3^{-x})^6\) is: \(h'(x) = 6(2^x + 3^{-x})^5 \cdot (2^x \ln(2) - 3^{-x} \ln(3))\).

Step 1: Inner function

The inner function is g(x) = 2^x + 3^(-x).

Step 2: Find the derivative of the inner function

To find the derivative, we use the power rule for exponential functions: \(\frac{d}{dx}a^x = a^x \ln(a)\). So, \(g'(x) = \frac{d}{dx} (2^x) + \frac{d}{dx} (3^{-x})\) \(g'(x) = 2^x \ln(2) - 3^{-x} \ln(3)\) (note the negative sign because of the derivative of -x) #Step 2: Identify the outer function and find its derivative#

Step 3: Outer function

The outer function is h(g(x)) = (g(x))^6.

Step 4: Find the derivative of the outer function

To find the derivative, we use the power rule: \(\frac{d}{dx}x^n = nx^{n-1}\). So, \(h'(g(x)) = \frac{d}{dx}((g(x))^6) = 6(g(x))^5\) #Step 3: Apply the chain rule to find the derivative of the original function#

Step 5: Apply the chain rule

According to the chain rule, the derivative of the original function h(x) is the product of the derivatives of the inner and outer functions: \(h'(x) = h'(g(x)) \cdot g'(x)\) \(h'(x) = 6(g(x))^5 \cdot (2^x \ln(2) - 3^{-x} \ln(3))\) #Step 4: Substitute the original inner function into the result#

Step 6: Substitute the original inner function

Now, replace g(x) with its original expression (2^x + 3^(-x)): \(h'(x) = 6(2^x + 3^{-x})^5 \cdot (2^x \ln(2) - 3^{-x} \ln(3))\) #Solution# The derivative of the function h(x) = (2^x + 3^(-x))^6 is: \(h'(x) = 6(2^x + 3^{-x})^5 \cdot (2^x \ln(2) - 3^{-x} \ln(3))\)

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that helps us find the derivative of composite functions.
When dealing with a function that contains other functions within it, like our example function, the derivative can be tricky to determine directly. This is where the chain rule comes in handy.Here’s how it generally works:

• Identify the outer and inner functions. In our exercise, the outer function is the power expression \((g(x))^6\), and the inner function is the combination of exponential terms \(g(x) = 2^x + 3^{-x}\).
• Take the derivative of each of these parts separately using their respective rules.
• Multiply these derivatives together. This product gives us the derivative of the composite function.

The application of the chain rule makes it possible to handle complex derivatives, as seen when we calculated \(h'(x) = 6(g(x))^5 \cdot g'(x)\). This equation combines the derivative of the outer function with the derivative of the inner function, showcasing the power of the chain rule.

Exponential Functions

Exponential functions are a key area in calculus, especially when it comes to differentiation.
These functions have the form \(a^x\), where \(a\) is a positive constant. The exponential increase or decrease is what makes them fascinating and crucial for various applications.To differentiate an exponential function, we utilize the formula:\[\frac{d}{dx}a^x = a^x \ln(a)\]This means that the derivative of an exponential function with base \(a\) is the function itself multiplied by the natural logarithm of the base (\(\ln(a)\)). In our exercise:

• We found the derivative of \(2^x\) to be \(2^x \ln(2)\).
• Similarly, the derivative of \(3^{-x}\) becomes \(-3^{-x} \ln(3)\) due to the negative exponent.

Having a good grasp of exponential functions and their derivatives is crucial for solving calculus problems effectively, making them easier to tackle in complex situations.

Power Rule

The power rule is another cornerstone principle in calculus, primarily used for finding derivatives of power functions. It's applicable to functions of the form \(x^n\), where \(n\) is any real number.
This rule is straightforward but powerful, making it very efficient for calculating derivatives.The general formula is:\[\frac{d}{dx}x^n = nx^{n-1}\]When applying the power rule in the context of composite functions, such as \((g(x))^6\) in our case, it becomes imperative to couple it with the chain rule.For the outer function in our example:

• The function is transformed from \((g(x))^n\) to produce the factor \(6(g(x))^5\).
• This is multiplied with the derivative of the inner function, resulting in a complete solution.

The power rule simplifies the differentiation process, especially when used alongside other rules such as the chain rule, as seen in how it helped solve the given problem efficiently.