Problem 77 - Brief Applied Calculus

Question

True or False: \(\frac{d}{d x} \sqrt{g(x)}=\sqrt{g(x)} \cdot g^{\prime}(x)\)

Step-by-Step Solution

Answer

False. The correct derivative is \( \frac{1}{2\sqrt{g(x)}} \cdot g'(x) \).

Step 1: Review the Derivative Rules

To solve this exercise, we need to use the chain rule for derivatives. The chain rule states that if you have a function composed with another function, i.e., \( f(g(x)) \), then the derivative \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \). For square root functions, recall that \( \sqrt{g(x)} \) can be written as \( (g(x))^{1/2} \).

Step 2: Differentiate the Square Root Function

Start by differentiating \( \sqrt{g(x)} = (g(x))^{1/2} \) using the chain rule. Let \( u = g(x) \), so \( \sqrt{g(x)} = u^{1/2} \). The derivative of \( u^{1/2} \) with respect to \( u \) is \( \frac{1}{2} u^{-1/2} = \frac{1}{2\sqrt{u}} \). Therefore, the derivative of \( \sqrt{g(x)} \) with respect to \( x \) is \( \frac{1}{2\sqrt{g(x)}} \cdot g'(x) \).

Step 3: Compare the Provided Expression with the Derivative

The provided expression is \( \sqrt{g(x)} \cdot g'(x) \), whereas the correct derivative we calculated is \( \frac{1}{2\sqrt{g(x)}} \cdot g'(x) \). Clearly, these two expressions are not the same.

Step 4: Conclusion

Given that the derivative \( \frac{d}{dx}\sqrt{g(x)} = \frac{1}{2\sqrt{g(x)}} \cdot g'(x) \) differs from the provided expression \( \sqrt{g(x)} \cdot g'(x) \), the statement is False.

Key Concepts

Understanding Derivative Rules

In calculus, derivative rules are essential for finding how a function changes. The derivative of a function, often denoted as \( f'(x) \), represents the rate of change or the slope of the function at any given point. Understanding these rules is key to solving many mathematical problems involving rates of changes, like speed and growth.
For basic functions, unique rules apply. For example, the power rule states that for a function of the form \( x^n \), the derivative is \( nx^{n-1} \). For more complex functions, combinations of functions require additional strategies, such as the product rule, quotient rule, and the chain rule.
Derivative rules are not just random formulas. They are logical results from the process of taking limits and understanding how a small change in \( x \) affects the corresponding change in \( y \). In the context of this exercise, the chain rule is used as it deals with a composite function \( f(g(x)) \). Make sure you are familiar with how each rule applies, as this will greatly enhance your problem-solving skills.

Square Root Functions in Calculus

Square root functions, represented as \( \sqrt{g(x)} \), can often seem intimidating because of their unique behavior compared to linear functions. However, they are simply power functions in disguise. By rewriting \( \sqrt{g(x)} \) as \( (g(x))^{1/2} \), the function becomes more accessible for applying derivative rules.
Square root functions have smooth curves, unlike the spiky behavior of other functions like polynomials. In calculus, differentiating square roots requires careful attention to the chain rule. This is because they often contain an inner function \( g(x) \) that complicates the differentiation slightly. It's important to remember that derivatives of square root functions will always result in a fraction, due to the negative exponent that appears when applying the power rule.
Understanding the behavior and transformation of square root expressions is invaluable, especially when you aim to differentiate functions involving square roots. Practice rewriting them in exponential form to make use of simpler differentiation rules.

Chain Rule Differentiation

The chain rule is a fundamental tool for differentiation in calculus. It allows you to differentiate complex compositions of functions by simplifying them into steps. The chain rule is written as follows: if \( h(x) = f(g(x)) \), then the derivative \( h'(x) = f'(g(x)) \cdot g'(x) \). It's like peeling an onion layer by layer, dealing with one function at a time.
To employ the chain rule effectively, identify the inner function \( g(x) \) and the outer function \( f(u) \), where \( u = g(x) \). First, differentiate the outer function with respect to \( u \), and afterwards, multiply by the derivative of the inner function with respect to \( x \).

• Identify the components: Recognize the outer and inner functions.
• Differentiate the outer function: Work with the function \( f(u) \) as if \( u \) is a simple variable.
• Multiply by the inner derivative: Account for the chain part by multiplying with \( g'(x) \).

In the exercise example provided, rewriting the square root as a power \( (g(x))^{1/2} \) makes it clear to apply the chain rule for accurate differentiation.