Problem 38 - Brief Applied Calculus

Question

31-38. Find the indicated derivatives. If \(f(x)=\frac{54}{\sqrt{x}}+12 \sqrt{x}\), find \(\left.\frac{d f}{d x}\right|_{x=9}\)

Step-by-Step Solution

Answer

The derivative of \( f(x) \) at \( x=9 \) is 1.

Step 1: Rewrite the Function

The given function is \( f(x) = \frac{54}{\sqrt{x}} + 12\sqrt{x} \). Rewrite it in terms of powers of \(x\). We know that \( \sqrt{x} = x^{1/2} \), so \( \frac{54}{\sqrt{x}} = 54x^{-1/2} \) and \( 12\sqrt{x} = 12x^{1/2} \). Thus, the function becomes \( f(x) = 54x^{-1/2} + 12x^{1/2} \).

Step 2: Differentiate the Function

Differentiate the rewritten function \( f(x) = 54x^{-1/2} + 12x^{1/2} \) with respect to \( x \). Use the power rule \( \frac{d}{dx} x^n = nx^{n-1} \). This gives:\[ \frac{df}{dx} = 54 \cdot (-\frac{1}{2})x^{-3/2} + 12 \cdot \frac{1}{2}x^{-1/2} \]Simplifying, we get:\[ \frac{df}{dx} = -27x^{-3/2} + 6x^{-1/2} \].

Step 3: Evaluate the Derivative at \(x=9\)

Substitute \( x = 9 \) into the derivative \( \frac{df}{dx} = -27x^{-3/2} + 6x^{-1/2} \) to find \( \left.\frac{df}{dx}\right|_{x=9} \):First, calculate each term:\[ -27(9^{-3/2}) = -27 \times \frac{1}{27} = -1 \]\[ 6(9^{-1/2}) = 6 \times \frac{1}{3} = 2 \]Adding these results gives:\[ -1 + 2 = 1 \].

Step 4: Conclude the Solution

The evaluated derivative at \( x = 9 \) is \( \left.\frac{d f}{d x}\right|_{x=9} = 1 \). Thus, the rate of change of \( f(x) \) at \( x = 9 \) is 1.

Key Concepts

power rule

The power rule is one of the simplest and most commonly used rules in calculus when dealing with derivatives. It states that if you have a function of the form \( x^n \), where \( n \) is any real number, the derivative (rate of change) of this function is \( n \times x^{n-1} \). This rule helps us to easily find the derivative of a function as long as it is expressed in terms of powers of \( x \).

For example, in the function \( f(x) = 54x^{-1/2} + 12x^{1/2} \), we apply the power rule as follows:

• For the term \( 54x^{-1/2} \), the derivative is \( \frac{d}{dx}(54x^{-1/2}) = 54 \cdot (-\frac{1}{2})x^{-3/2} \).
• For the term \( 12x^{1/2} \), the derivative is \( \frac{d}{dx}(12x^{1/2}) = 12 \cdot \frac{1}{2}x^{-1/2} \).

The simplicity of the power rule makes it extremely useful, especially when dealing with more complex functions involving different techniques of differentiation.

differentiation techniques

Differentiation techniques are essential tools in calculus, which allow us to find the derivative or the rate of change of a function with respect to a variable. There are several key techniques available, but here, we focus on the power rule as it pertains to polynomial functions. This technique is incredibly efficient for terms expressed as powers of \( x \).

In the exercise, rewriting the function \( f(x) = 54x^{-1/2} + 12x^{1/2} \) allows easier application of differentiation techniques. By converting square roots and division into exponential form:

• \( \frac{54}{\sqrt{x}} \) becomes \( 54x^{-1/2} \).
• \( 12\sqrt{x} \) becomes \( 12x^{1/2} \).

Once rewritten, employing the power rule—a core differentiation technique—becomes straightforward. Simply take each term, multiply by the exponent, and reduce the exponent by one. This unified approach is vital in handling various forms, simplifying the process, and arriving at the correct derivative rapidly.

rate of change

The rate of change of a function at a particular point is an important concept in calculus. It indicates how much the function value changes as the input changes slightly. In practical terms, the derivative of a function at a point gives its rate of change.

In the exercise, we find the rate of change of the function \( f(x) = \frac{54}{\sqrt{x}} + 12 \sqrt{x} \) at \( x = 9 \). By differentiating the function, we determine \( \frac{df}{dx} = -27x^{-3/2} + 6x^{-1/2} \). Evaluating this expression at \( x = 9 \) involves substituting and simplifying as follows:

• For \( -27x^{-3/2} \), substitute \( x=9 \) to get \(-1\).
• For \( 6x^{-1/2} \), substitute \( x=9 \) to get \(2\).

Adding these gives \( 1 \), indicating the rate of change of the function is \( 1 \) at \( x = 9 \). It means for a slight increase in \( x \), \( f(x) \) increases by 1 unit, providing insight into the function's behavior at that point.