Problem 4 - Calculus and its applications

Question

Differentiate each function $$ y=(8-x)^{100} $$

Step-by-Step Solution

Answer

The derivative of \( y =(8-x)^{100} \) is \( -100(8-x)^{99} \).

Step 1: Apply the Power Rule

The function is a power function in the form of \( y = (u(x))^n \), where \( u(x) = 8-x \) and \( n = 100 \). The power rule for differentiation states that \( \frac{d}{dx}[u(x)]^n = n \cdot [u(x)]^{n-1} \cdot u'(x) \).

Step 2: Differentiate the Inner Function

Next, differentiate the inner function \( u(x) = 8-x \) with respect to \( x \). The derivative is \( u'(x) = \frac{d}{dx}(8-x) = 0 - 1 = -1 \).

Step 3: Combine the Results

Substitute \( n = 100 \), \( [u(x)]^{n-1} = (8-x)^{99} \), and \( u'(x) = -1 \) into the formula found in Step 1. Thus, the derivative \( \frac{dy}{dx} = 100 \cdot (8-x)^{99} \cdot (-1) \).

Step 4: Simplify the Expression

Finally, simplify the expression: \( 100 \cdot (8-x)^{99} \cdot (-1) = -100 (8-x)^{99} \).

Key Concepts

Power Rule

The Power Rule is a crucial concept in differentiation within calculus. It simplifies finding the derivative of functions where the variable is raised to a power. For example, if you have a function of the form \( y = x^n \), the derivative is calculated as \( \frac{dy}{dx} = n \cdot x^{n-1} \).
This rule makes it easy because you bring down the exponent as a coefficient and then reduce the exponent by one. In the original exercise, we had \( y = (8-x)^{100} \). Here, the base \( u(x) = 8-x \) is not just a simple \( x \) but a complete expression. Applying the power rule involves more than just exponent manipulation. You must also consider the differential of the inner expression, which leads us to the Chain Rule.

Chain Rule

The Chain Rule is a powerful tool used to differentiate composite functions, where one function is nested within another. When you have a function like \( y = (u(x))^n \), you need to differentiate both the outer function and the inner function. The formula for the Chain Rule is:

• \( \frac{d}{dx}[u(x)]^n = n \cdot [u(x)]^{n-1} \cdot u'(x) \)

In our exercise, the function is \( y = (8-x)^{100} \). Here, the outer function is \( (8-x)^{100} \) with an exponent of 100, and the inner function is \( u(x) = 8-x \). To apply the Chain Rule:- Compute the derivative of the outer function by applying the Power Rule: multiply by the exponent and reduce it by one, resulting in \( 100 \cdot (8-x)^{99} \).- Then find the derivative of the inner function, which is simply \( -1 \), since the derivative of \( 8-x \) is \( 0-1 \).- Finally, multiply these two results together to get the derivative.

Calculus Problem Solving

Solving calculus problems efficiently relies on understanding and applying rules logically and methodically. It involves breaking down complex expressions into manageable parts using differentiation rules like the Power and Chain Rules.To solve a differentiation problem successfully:

• Identify the type of function you are dealing with (e.g., power, polynomial, trigonometric).
• Determine which differentiation rules apply, such as the Power Rule for simple exponents or the Chain Rule for nested functions.
• Carefully calculate derivatives, ensuring you follow the order of operations and combine results accurately.
• Simplify the derivative if possible, as seen in our exercise where \( -100(8-x)^{99} \) was the final simplified derivative.

Approaching calculus problems with these steps helps you avoid errors and develop a deeper understanding of the mathematical concepts involved. Each problem enhances your ability to tackle new and more complex equations by building a strong foundation in applying these essential calculus techniques.