Problem 22 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(h(x)=\sqrt[3]{1-2 x}\)

Step-by-Step Solution

Answer

The derivative is \( h'(x) = -\frac{2}{3}(1-2x)^{-\frac{2}{3}} \).

Step 1: Rewrite the Function in Exponential Form

The given function is written as a cube root: \( h(x) = \sqrt[3]{1 - 2x} \). To make differentiation easier, we rewrite this expression using exponents: \( h(x) = (1 - 2x)^{\frac{1}{3}} \).

Step 2: Apply the Chain Rule

The Chain Rule is necessary to differentiate composite functions. Here, \( h(x) = g(f(x)) \) with \( g(u) = u^{\frac{1}{3}} \) and \( f(x) = 1-2x \). According to the Chain Rule, the derivative \( h'(x) = g'(f(x)) \cdot f'(x) \).

Step 3: Differentiate the Outer Function

First, differentiate \( g(u) = u^{\frac{1}{3}} \) with respect to \( u \). Using the power rule, we get \( g'(u) = \frac{1}{3}u^{\frac{1}{3} - 1} = \frac{1}{3}u^{-\frac{2}{3}} \).

Step 4: Differentiate the Inner Function

Next, differentiate \( f(x) = 1-2x \) with respect to \( x \). The derivative is \( f'(x) = -2 \) since the derivative of a constant is zero and the derivative of \(-2x\) is \(-2\).

Step 5: Combine the Results Using the Chain Rule

Substitute the derivatives into the Chain Rule formula: \( h'(x) = g'(f(x)) \cdot f'(x) \). This gives us \( h'(x) = \frac{1}{3}(1-2x)^{-\frac{2}{3}} \cdot (-2) \).

Step 6: Simplify the Expression

Simplify the expression for \( h'(x) \): \[ h'(x) = -\frac{2}{3}(1-2x)^{-\frac{2}{3}} \].

Key Concepts

Chain Rule in Calculus Differentiation

The Chain Rule is a fundamental concept in calculus used when differentiating composite functions—functions that are formed by applying one function to the results of another. It helps us tackle situations where a straightforward application of differentiation rules is insufficient. For the function

• \( h(x) = \sqrt[3]{1 - 2x} \)
• Which becomes \( (1 - 2x)^{\frac{1}{3}} \)

The Chain Rule allows us to differentiate by expressing \( h(x) \) as \( g(f(x)) \), where:

• Outer function \( g(u) = u^{\frac{1}{3}} \)
• Inner function \( f(x) = 1 - 2x \)

Through the Chain Rule formula, \( h'(x) = g'(f(x)) \cdot f'(x) \), we first find the derivative of the outer function with respect to its variable, and then multiply by the derivative of the inner function with respect to \( x \). This method ensures we account for the function's layered construction, giving an accurate representation of its rate of change.

Understanding Exponential Form

Exponential forms provide a useful means of rewriting root expressions to facilitate differentiation using basic derivative rules. In the given exercise, the cube root \( \sqrt[3]{1 - 2x} \) is expressed as an exponent, \((1 - 2x)^{\frac{1}{3}}\).

This conversion applies the property where any root \( \sqrt[n]{a} \) is equivalent to \( a^{\frac{1}{n}} \). By rewriting roots as exponents, we can apply the power rule more easily: the derivative of a power function \( u^n \) is \( nu^{n-1} \). This technique is essential as it transforms complex root-based expressions into a familiar format, paving the way for simple and efficient differentiation.

Composite Functions and Their Derivatives

Composite functions are combinations of two or more functions where the output of one function becomes the input of another. They are common in advanced mathematics and require careful manipulation using specific rules to differentiate accurately. In our exercise, the function \( h(x) = \sqrt[3]{1-2x} \) can be viewed as:

• \( g(f(x)) \) with outer function \( g(u) = u^{\frac{1}{3}} \)
• Inner function \( f(x) = 1-2x \)

To differentiate, we treat these as distinct steps:

• Differentiate the outer function \( g(u) \) first.
• Follow by differentiating the inner function \( f(x) \).

Subsequently, we combine these using the Chain Rule. This process of breaking down each function part ensures we accurately derive complex expressions that depend on another's output. Mastery of composite functions is key to understanding and solving higher-level calculus problems efficiently.