Problem 42 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(37-58\) , find the derivative of the function. (Hint: In some exercises, you may find it helpful to apply logarithmic properties before differentiating.) $$ y=x\left(6^{-2 x}\right) $$

Step-by-Step Solution

Answer

The derivative of the function \(y = x\left(6^{-2 x}\right)\) with respect to x is \(y' = 6^{-2x}(-2x\ln(6) + 1)\)

Step 1: Understand the function

The given function is \(y=x\left(6^{-2 x}\right)\). This is a product of \(x\) and \(6^{-2x}\). The derivative of this function can be found using the product rule.

Step 2: Apply the Product Rule

The product rule states that the derivative of the product of two functions is derivative of the first times the second plus the first times the derivative of the second. So let's set \(u = x\) and \(v = 6^{-2x}\). Now the derivative of the function will be \(u'v + uv'\).

Step 3: Find the derivatives u' and v'

Derivative of \(u = x\) with respect to x is \(u' = 1\). The derivative of \(v = 6^{-2x}\) requires use of chain rule, and it will be: \(v' = -2 \cdot 6^{-2x} \cdot \ln(6)\).

Step 4: Substitute into the Product Rule

Substitute \(u\), \(v\), \(u'\), and \(v'\) into the formula from Step 2: \(y' = u'v + uv' = x \cdot -2 \cdot 6^{-2x} \cdot \ln(6) + 1 \cdot 6^{-2x}\)

Step 5: Simplify the expression

We can simplify \(y'\) by factoring out common elements, \(6^{-2x}\): \(y' = 6^{-2x}(-2x\ln(6) + 1)\)

Key Concepts

Product Rule

In calculus, the product rule is a fundamental tool when taking derivatives of functions multiplied together. It is particularly useful when you encounter two differentiable functions in a product. To apply the product rule, if you have a function that's the product of two functions, say \( u(x) \) and \( v(x) \), the derivative is given by:

• Derivative of the first function, multiplied by the second function.
• Plus, the first function multiplied by the derivative of the second function.

In mathematical terms, the formula is:\[ (uv)' = u'v + uv' \]This rule helps you find the derivative of a product of functions without having to expand them. It's a time-saver and essential for working with complex expressions.

Chain Rule

The chain rule is another cornerstone in calculus, especially useful when dealing with composite functions. A composite function occurs when you "nest" functions, meaning one function is within another. To differentiate such functions, the chain rule provides a method.

• First, it involves taking the derivative of the outer function.
• Then, multiply it by the derivative of the inner function.

For a function \( f(g(x)) \), its derivative is:\[ f'(g(x)) \, g'(x) \]Applying the chain rule ensures that you account for all the changes in the nested structures. In our original problem, this rule was applied when differentiating \( 6^{-2x} \).

Logarithmic Differentiation

Logarithmic differentiation is an elegant technique useful for differentiating functions more complicated in form, particularly those involving products, quotients, or powers. This method involves taking the natural logarithm of both sides of an equation, which simplifies the process of differentiation.

• Take the natural log of both sides of the function.
• Use logarithmic properties to simplify if possible.
• Differentiate both sides.
• Finally, solve back for the original function’s derivative.

When a function poses challenges due to its structure, applying logarithmic differentiation can transform the problem into something more manageable. It's a powerful technique when faced with functions like exponentials and powers.

Function Differentiation

Function differentiation involves calculating the derivative, which represents the rate at which one quantity changes with respect to another. It is the essence of calculus and is crucial for analyzing and understanding the behavior of functions.

• Helps determine instantaneous rate of change.
• Useful for finding slopes of tangent lines.
• Essential in optimizing problems, like finding maxima and minima.

Whether you're working on simple functions or more advanced compositions, mastering differentiation allows for better insights into how functions behave. In the original exercise, understanding how to differentiate the given function thoroughly relied on combining several differentiation rules effectively.