Problem 47 - Calculus and its applications

Question

Find \(f^{\prime}(x)\) for the given function. \(f(x)=\frac{1}{\sqrt{x}}(\) See Exercise 56 in Section \(1.3 .)\)

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Function

The given function is \( f(x) = \frac{1}{\sqrt{x}} \). To find its derivative, it's helpful to rewrite it using exponent rules. We know that \( \sqrt{x} = x^{1/2} \). Thus, the function becomes \( f(x) = x^{-1/2} \).

Step 2: Apply the Power Rule

The power rule for differentiation states that if \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \). Here, \( n = -1/2 \). Applying the power rule gives \( f'(x) = -\frac{1}{2} \cdot x^{-1/2 - 1} \).

Step 3: Simplify the Derivative

Simplify the expression for the derivative. We have \( f'(x) = -\frac{1}{2} \cdot x^{-3/2} \). This can be rewritten with the negative exponent as \( f'(x) = -\frac{1}{2x^{3/2}} \).

Key Concepts

Power Rule

The power rule is a fundamental concept in calculus used to differentiate functions of the form \( f(x) = x^n \). It provides a quick and efficient way to find derivatives. The rule states that when you have a power function, the derivative is calculated by multiplying the exponent by the function and then decreasing the exponent by one. In formulaic terms, it can be expressed as:

• If \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \)

This rule is particularly useful because it simplifies the process of differentiation for any polynomial term. For example, if you have \( f(x) = x^{4} \), applying the power rule gives you \( f'(x) = 4 \cdot x^{3} \). It's a simple shortcut that helps avoid more complex calculations and is one of the first tools students learn when tackling calculus problems. You only need to know the original exponent, and you can immediately find the derivative.

Differentiation

Differentiation is the process of finding the derivative of a function. It measures how a function's output changes as its input changes, essentially describing the function's rate of change. In a graphical sense, differentiation helps determine the slope of a tangent line to a curve at any given point.

• This process is essential in calculus and is applied across various fields such as physics, economics, and biology.
• By differentiating, we can find crucial points like maxima, minima, and inflection points, which are pivotal for graph analysis and optimization problems.

The derivative of a function at any point gives the instantaneous rate of change, providing valuable insights into the function's behavior. With differentiation, you can explore the intricate dynamics of curves and surface contours, making it a powerful tool for mathematical analysis.

Exponent Rules

Exponent rules are mathematical guidelines for simplifying expressions involving powers of numbers or variables. They are crucial when rewriting functions, as they help convert complex expressions into simplified forms that are more manageable for operations like differentiation.

• The product rule for exponents states that \( x^a \times x^b = x^{a+b} \).
• The quotient rule provides that \( \frac{x^a}{x^b} = x^{a-b} \).
• The power rule for exponents indicates that \( (x^a)^b = x^{a\cdot b} \).
• Any base raised to the zero power is equal to one: \( x^0 = 1 \).

In the context of the given exercise, using exponent rules allowed the transformation from \( \frac{1}{\sqrt{x}} \) to \( x^{-1/2} \). This transformation is necessary to make the function compliant with the power rule for differentiation. Simplifying complex terms using these rules often makes the subsequent application of calculus principles much more straightforward.