Problem 20 - The Calculus with Analytic Geometry

Question

Find the derivative of the given function. $$ F(x)=\frac{(5 x-8)^{-2}}{\left(x^{2}+3\right)^{-3}} $$

Step-by-Step Solution

Answer

The derivative is \( F'(x) = -10(5x - 8)^{-3} (x^2 + 3)^3 + 6x(5x - 8)^{-2} (x^2 + 3)^2 \).

Step 1: Rewrite the Function in Simpler Form

First, rewrite the given function in a simpler form suitable for differentiation. Given: \[ F(x) = \frac{(5x - 8)^{-2}}{(x^2 + 3)^{-3}} \] Rewrite it as: \[ F(x) = (5x - 8)^{-2} \cdot (x^2 + 3)^3 \]

Step 2: Apply the Product Rule

To differentiate \( F(x) \), apply the product rule: If \( F(x) = u \cdot v \), then \( F'(x) = u'v + uv' \). Here, let: \( u = (5x - 8)^{-2} \) and \( v = (x^2 + 3)^3 \).

Step 3: Differentiate Each Function Separately

Compute the derivatives of \( u \) and \( v \) separately. For \( u = (5x - 8)^{-2} \): \[ u' = -2(5x - 8)^{-3} \cdot 5 = -10(5x - 8)^{-3} \] For \( v = (x^2 + 3)^3 \): \[ v' = 3(x^2 + 3)^2 \cdot 2x = 6x(x^2 + 3)^2 \]

Step 4: Use the Product Rule Formula

Substitute the derivatives back into the product rule formula: \[ F'(x) = u'v + uv' \] Now, substitute \( u' \), \( u \), \( v \), and \( v' \): \[ F'(x) = -10(5x - 8)^{-3} \cdot (x^2 + 3)^3 + (5x - 8)^{-2} \cdot 6x(x^2 + 3)^2 \]

Step 5: Simplify the Expression

Simplify the expression obtained in the previous step: Combine the two terms into a single fraction (if possible) and simplify. This is the derivative of the given function.

Key Concepts

Product Rule

The Product Rule is a fundamental technique in calculus used to find the derivative of the product of two functions. The rule states:

• If you have a function \( F(x) = u(x) \times v(x) \), the derivative \( F'(x) \) can be found using: \( F'(x) = u'(x) v(x) + u(x) v'(x) \).

Breaking this down:

• First, differentiate each function individually (i.e., find \( u'(x) \) and \( v'(x) \)).
• Then, multiply the derivative of the first function by the second function and add it to the product of the first function and the derivative of the second function.

For our exercise, let \( u = (5x - 8)^{-2} \) and \( v = (x^2 + 3)^3 \). We found \( u' \) and \( v' \) separately and then combined them using the product rule formula to find \( F'(x) \). Keeping these steps in mind helps when working with any problems involving the product of two functions.

Power Rule

The Power Rule is another essential tool in differentiation. This rule applies whenever you have a function of the form \( f(x) = x^n \), where \( n \) is any real number. The Power Rule states:

• If \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \).

This means you bring the exponent down as a coefficient in front and subtract one from the exponent. This rule simplifies the differentiation process significantly.
For our exercise:

• We applied the Power Rule to differentiate \( u = (5x - 8)^{-2} \).
• This gave us \( u' = -2(5x - 8)^{-3} \) after applying the chain rule to account for the inner function \( 5x - 8 \).
• Similarly, for \( v = (x^2 + 3)^3 \), applying the Power Rule yielded \( v' = 6x(x^2 + 3)^2 \) through the chain rule, adjusting for the inner function \( x^2 + 3 \).

Understanding the Power Rule makes it easier to handle polynomial and rational functions quickly.

Function Differentiation

Function differentiation involves finding the derivative of functions to understand their rates of change. Differentiation is a core concept in calculus, deeply interconnected with limits.
In our specific exercise, we focused on differentiating a composite function that required combining multiple differentiation rules:

• First, we rewrote the function for simplification, transforming our original fraction into a product of two functions.
• Then, we applied the Product Rule, identifying suitable \( u \) and \( v \) components.
• Next, differentiating each component using the Power Rule and Chain Rule simultaneously.
• We finally combined the differentiated components to get the overall derivative of the function.

Having a clear grasp of these differentiation techniques is essential for solving more complex calculus problems. Combining the rules proficiently allows for seamless function differentiation, integral to understanding function behavior and change over time.