Question

Derivatives Find the derivative of the following functions. See Example 2 of Section 3.2 for the derivative of \(\sqrt{x}\). $$f(s)=\frac{\sqrt{s}}{4}$$

Short answer

Answer: The derivative of the function \(f(s)=\frac{\sqrt{s}}{4}\) with respect to \(s\) is \(f'(s)=0\).

Step-by-step solution

Rewrite the function as a power

The function \(f(s)\) can be rewritten as $$f(s)=\frac{s^{\frac{1}{2}}}{4}$$ where \(\sqrt{s}\) has been replaced with \(s^{\frac{1}{2}}\).

Apply the power rule

The power rule states that the derivative of \(s^n\) is \(ns^{n-1}\), where n is any constant. In this case, we have $$f'(s)=\frac{1}{2}s^{\frac{1}{2}-1}\cdot\frac{d}{ds}\left(\frac{1}{4}\right)$$

Simplify the expression

Simplify the expression for \(f'(s)\): $$f'(s)=\frac{1}{2}s^{\frac{-1}{2}}\cdot\frac{d}{ds}\left(\frac{1}{4}\right)$$ Since the derivative of any constant, such as \(\frac{1}{4}\), with respect to a variable is 0, we have $$f'(s)=\frac{1}{2}s^{\frac{-1}{2}}\cdot 0 = 0$$ Therefore, the derivative of the function \(f(s)=\frac{\sqrt{s}}{4}\) is: $$f'(s)=0$$

Key concepts

Understanding the Power Rule

Taking derivatives is a fundamental concept in calculus, and the power rule is one of the most commonly used techniques. The power rule simplifies the process of differentiation when dealing with powers of a variable.

Let's consider you have a function of the form \( f(x) = x^n \), where \( n \) is any real number. If you apply the power rule for differentiation, the derivative of the function \( f(x) \) with respect to \( x \) would be \( f'(x) = nx^{n-1} \).

For example, if the function is \( g(x) = x^3 \), then the derivative using the power rule is \( g'(x) = 3x^{3-1} = 3x^2 \). Remember, the power rule lets you bring down the exponent as a coefficient in front of the variable and then subtract one from the original exponent to find the new power of the variable in the derivative.

Derivative of a Constant

When you are differentiating functions, you will often encounter constants. A key concept to grasp when finding derivatives is the derivative of a constant. Simply put, the derivative of any constant with respect to any variable is zero.

Why is this the case? Well, intuitively, a constant number does not change; it has no rate of change, which is what a derivative measures. For example, if \( h(x) = 5 \), a constant function, its derivative is \( h'(x) = 0 \). This concept applies when the constant is part of a larger function, like \( f(s) = \frac{\sqrt{s}}{4} \). The \( \frac{1}{4} \) part is constant, so its derivative does not contribute to the overall derivative of the function.

Simplifying Expressions in Calculus

In calculus, simplifying expressions is crucial to understanding and solving problems effectively. Simplification involves breaking down complex expressions into simpler forms that are easier to work with. When simplifying expressions to find derivatives, remember to combine like terms, cancel out terms where possible, and use basic algebraic rules.

For example, if you found yourself with a derivative expression like \( f'(s) = 2s^{-1} + 3s^{0} - 4 \), you could simplify this expression by recognizing that any variable to the power of zero is one (\( s^{0} = 1 \)), and that a negative exponent indicates division by that variable (\( s^{-1} = \frac{1}{s} \)). Thus, the simplified form of the derivative would be \( f'(s) = 2/s + 3 - 4 \). Always simplify the expressions to their least complex form to make further calculations or interpretations of the derivative more manageable.