Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y / d x\). \(y=(3 x-9)^{-5 / 3}\)

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -5(3x-9)^{-\frac{8}{3}} \)

Step 1: Identify the Use of the Chain Rule

We notice that the given function is of the form \( (u(x))^n \) where \( u(x) = 3x - 9 \) and \( n = -\frac{5}{3} \). We need to apply the chain rule for differentiation, which states that \( \frac{d}{dx}(u(x))^n = n(u(x))^{n-1} \cdot \frac{du}{dx} \).

Step 2: Differentiate the Outer Function

Differentiate the outer function \( (u(x))^n \) with respect to \( u \). The derivative is \( n(u(x))^{n-1} \), so here it becomes \( -\frac{5}{3}(3x-9)^{-\frac{5}{3} - 1} = -\frac{5}{3}(3x-9)^{-\frac{8}{3}} \).

Step 3: Differentiate the Inner Function

Next, differentiate the inner function \( u(x) = 3x - 9 \) with respect to \( x \). The derivative is \( \frac{du}{dx} = 3 \).

Step 4: Apply the Chain Rule

Combine both derivatives according to the chain rule. Multiply the derivative of the outer function by the derivative of the inner function: \( \frac{dy}{dx} = -\frac{5}{3}(3x-9)^{-\frac{8}{3}} \times 3 \).

Step 5: Simplify the Expression

Finally, simplify the expression: \( \frac{dy}{dx} = -5(3x-9)^{-\frac{8}{3}} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. In simpler terms, it is used to find the derivative, which gives us a mathematical description of how a function's output value changes as its input value changes. Differentiation is essential for analyzing graphs, determining slopes of curves, and solving real-world problems such as motion and growth.

Key aspects of differentiation include:

• Identifying the function you want to differentiate.
• Applying the correct differentiation rules, such as product rule, quotient rule, or chain rule, based on the function's form.
• Understanding that differentiation requires careful handling of algebraic expressions to achieve accurate results.

In the given exercise, differentiation is used to identify the change rate of the function \((3x-9)^{-5/3}\). This requires understanding the nested nature of functions and applying the chain rule to manage this complexity.

Calculus

Calculus is a branch of mathematics that studies continuous change. It is divided into two main areas: differential calculus and integral calculus. Differential calculus focuses on the concept of the derivative, which measures how a quantity changes in response to changes in another quantity.

In practical terms, calculus allows us to describe how things change over time and space. It's widely used in a variety of fields such as physics, engineering, economics, and biology. To successfully apply calculus, one needs a strong understanding of functions, limits, and an ability to perform algebraic manipulations.

The given exercise focuses on using differential calculus to determine the rate of change of a power function, involving careful differentiation through techniques like the chain rule.

Derivative of Power Function

The derivative of a power function is an expression that shows how a function involving exponents changes as its input changes. A power function is typically of the form \(u(x)^n\), where \(u(x)\) is a differentiable function and \(n\) is a constant exponent. The work for finding the derivative of a power function is entrenched in the rules of differentiation, particularly when you deal with the Chain Rule.

For the derivative of \(u(x)^n\):

• Apply the power rule: Bring down the exponent \(n\) and reduce the original exponent by one, i.e., \(n(u(x))^{n-1}\).
• Multiply this result by the derivative of the inside function \(u(x)\).

In our exercise, the function \(y = (3x-9)^{-5/3}\) requires these steps, identifying \(u(x) = 3x - 9\) and using the power rule combined with the chain rule to find the derivative correctly. As a result, you can derive the pattern that shows how the function's gradient alters at different points.