Problem 28 - The Calculus with Analytic Geometry

Question

Find the derivative of the given function. (HIN\Upsilon: \(\left.|a|=\sqrt{a^{2}} .\right)\) $$ h(x)=\sqrt[3]{|x|+x} $$

Step-by-Step Solution

Answer

For \(x > 0\), \( h'(x) = \frac{2^{1/3}}{3} x^{-2/3} \. For \(x \leq 0\), \ h'(x) = 0 \.

Step 1: Rewrite the Absolute Expression

First, recognize that \(|x|\) can be rewritten as \(\text{sgn}(x) x\), where \(\text{sgn}(x)\) is the sign function. Therefore, \(|x| + x\) can be written as \(\text{sgn}(x) x + x = (1 + \text{sgn}(x)) x\).

Step 2: Update the Original Function

Rewrite the function \(h(x)\) as \(\big( (1 + \text{sgn}(x)) x \big)^{1 / 3}\). This uses the fact that the argument inside the cube root can be simplified.

Step 3: Apply the Sign Function

Since the sign function \(\text{sgn}(x)\) is either 1, -1, or 0, we can split the function into two cases: \(x eq 0\) and \(x = 0\). When \(x > 0\), \(\text{sgn}(x) = 1\), making the function \(h(x) = (2x)^{1 / 3}\). When \(x < 0\), \(\text{sgn}(x) = -1\), making the function \(h(x) = 0\).

Step 4: Differentiate for Each Case

For the case \(x > 0\), differentiate \(h(x) = (2x)^{1 / 3}\) using the chain rule. Let \(u = 2x\), then \(h(x) = u^{1 / 3}\). The derivative is \( (1/3) u^{-2 / 3} \frac{du}{dx} = (1/3) (2x)^{-2 / 3} \times 2 \), which simplifies to \( \frac{2}{3} \times (2x)^{-2/3} \). For the case \(x < 0\), \(h(x) = 0\) so \(h'(x) = 0\). Thus, for \(x = 0\), similarly use the continuity argument for derivative.

Step 5: Combine the Results

Combine the differentiated results. For \(x > 0\), the derivative is \(h'(x) = \frac{2}{3} \times (2x)^{-2/3} = \frac{2}{3} \times 2^{-2/3} x^{-2/3} = \frac{2^{1/3}}{3} x^{-2/3}\). For \(x < 0\), the derivative is \(h(x) = 0\).

Key Concepts

Derivative Rules

To find derivatives efficiently, it is crucial to understand the various derivative rules. These rules help us simplify and find the derivatives of complex functions. Some of the fundamental rules are:

• Power Rule: If \(f(x) = x^n\), then the derivative \(f'(x) = nx^{n-1}\).
• Product Rule: If \(f(x) = u(x) \times v(x)\), then the derivative \(f'(x) = u'(x)v(x) + u(x)v'(x)\).
• Quotient Rule: If \(f(x) = \frac{u(x)}{v(x)}\), then the derivative \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\).
• Chain Rule: If \(f(x) = g(h(x))\), then the derivative \(f'(x) = g'(h(x)) \times h'(x)\).

By mastering these rules, you can tackle more challenging calculus problems with confidence.

Absolute Value Function

The absolute value function, denoted as \(|x|\), represents the distance of a number from zero on the number line. It is always non-negative. Mathematically, it is defined as:
\[|x| = \begin{cases} x, & \text{if } x \geq 0\ -x, & \text{if } x < 0 \end{cases}\]
For example, \(|3| = 3\) and \(|-3| = 3\). This property makes the absolute value function piecewise, meaning it is defined by different expressions based on the value of \(x\). In our exercise, we use the hint that \(|a| = \sqrt{a^2}\) to simplify calculations involving absolute values.

Piecewise Functions

Piecewise functions are functions defined by multiple sub-functions, each applying to a specific interval of the function’s domain. These are particularly useful when dealing with cases where different rules apply in different contexts. For instance, the absolute value function is a classic example:
\[|x| = \begin{cases} x, & \text{if } x \geq 0\ -x, & \text{if } x < 0 \end{cases}\]
In the given exercise, the function splits into cases based on the sign of \(x\). It is essential to handle each case separately to find the correct derivative.

Chain Rule in Calculus

The chain rule is a fundamental tool in calculus for finding the derivative of composite functions. If you have a function \(f(x) = g(h(x))\), the chain rule states that the derivative is:
\[f'(x) = g'(h(x)) \times h'(x)\]
This rule allows you to differentiate complex functions by breaking them down into simpler components. For our exercise, we use the chain rule when differentiating \(h(x) = (2x)^{1/3}\). By letting \(u = 2x\), we differentiate \(u^{1/3}\) and then multiply by the derivative of \(u\). This method is powerful for tackling nested functions with ease.