Problem 24 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$q=\sqrt[3]{2 r-r^{2}}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{2(1 - r)}{3(2r - r^2)^{2/3}} \).

Step 1: Understand the Function

The function given is a cube root: \(q = \sqrt[3]{2r - r^2}\). To find its derivative, we will use the chain rule and the power rule.

Step 2: Rewrite in Power Form

Rewrite the cube root in terms of an exponent: \(q = (2r - r^2)^{1/3}\). This makes it easier to apply the differentiation rules.

Step 3: Apply the Chain Rule

To differentiate \(q = (2r - r^2)^{1/3}\), apply the chain rule, which gives \( \frac{dq}{dr} = \frac{1}{3}(2r - r^2)^{-2/3}\cdot (\text{derivative of } 2r - r^2) \).

Step 4: Differentiate the Inner Function

Find the derivative of the inner function \(2r - r^2\). Using basic differentiation rules, this is \(2 - 2r\).

Step 5: Combine the Derivatives

Substitute the derivative of the inner function back into the chain rule expression: \(\frac{dq}{dr} = \frac{1}{3}(2r - r^2)^{-2/3} \cdot (2 - 2r)\).

Step 6: Simplify the Expression

Simplify the expression: \(\frac{dq}{dr} = \frac{2(1 - r)}{3(2r - r^2)^{2/3}}\). This is the simplified form of the derivative.

Key Concepts

Chain Rule

The Chain Rule is an essential concept in calculus used for differentiating composite functions, which are functions within functions. In essence, if you have a situation where one function is nested inside another, like \( q = f(g(r)) \), the Chain Rule helps us find the derivative efficiently.

The Chain Rule states:

• If \( y = f(u) \) and \( u = g(x) \), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

When applying the Chain Rule to a function like \( q = (2r - r^2)^{1/3} \), first, identify the outer and inner functions. Here:

• Outer function (\( f(u) \)): \( u^{1/3} \).
• Inner function (\( g(r) \)): \( 2r - r^2 \).

Differentiate the outer function with respect to the inner function, and multiply it by the derivative of the inner function with respect to \( r \). This process allows us to manage complex derivations step-by-step.

Power Rule

The Power Rule is one of the most straightforward rules in calculus, used to find the derivative of functions in the form \( ax^n \), where \( a \) is a constant and \( n \) is any real number.

The Power Rule states:

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

In the context of cube roots, like \( q = (2r - r^2)^{1/3} \), the Power Rule is applied to derive \( q \) with respect to \( r \).

In this case, recognize that the exponent \( 1/3 \) is the \( n \) in the Power Rule. Therefore, when applying the Power Rule to the rewritten power form \( q = (2r - r^2)^{1/3} \):

• The derivative is \( \frac{1}{3}(2r - r^2)^{-2/3} \), calculated by multiplying the exponent by the same expression raised to the power of \( n - 1 \).

Understanding this rule simplifies the process of finding derivatives for functions involving powers.

Cube Root Differentiation

Cube Root Differentiation refers to the differentiation of functions involving cube roots, an important process especially when dealing with polynomial expressions under the root. A cube root function follows the form \( q = \sqrt[3]{x} \), which can be tricky at first fly but becomes manageable by rewriting it using exponents.

For example, the function \( q = \sqrt[3]{2r - r^2} \) can be rewritten into \( q = (2r - r^2)^{1/3} \). This new form allows us to apply both the Chain Rule and the Power Rule.

Steps to Differentiating Cube Root Functions:

• Rewrite the Function: Convert cube roots into exponential form such as \( x^{1/3} \).
• Apply Differentiation Rules: Use the Power Rule and, if necessary, the Chain Rule, to find derivatives efficiently.
• Simplify Your Derivative: Once differentiated, simplify the derivative expression as much as possible for clarity and ease of understanding.

These strategies make it easier to handle differentiation when dealing with cube root functions, enhancing your calculus toolkit.