Question

(a) Use the Quotient Rule to differentiate the function. (b) Manipulate the function algebraically and differentiate without the Quotient Rule. (c) Show that the answers from \((\mathrm{a})\) and \((\mathrm{b})\) are equivalent. $$h(s)=\frac{3}{4 s^{3}}$$

Short answer

Both methods give \( h'(s) = \frac{-9}{4s^4} \).

Step-by-step solution

Differentiate using the Quotient Rule

The Quotient Rule states that for a function \( h(s) = \frac{u(s)}{v(s)} \), the derivative \( h'(s) \) is \[ h'(s) = \frac{u'(s)v(s)-u(s)v'(s)}{[v(s)]^2} \].For \( h(s) = \frac{3}{4s^3} \), we assign \( u(s) = 3 \) (constant) and \( v(s) = 4s^3 \). Thus, \( u'(s) = 0 \) and \( v'(s) = 12s^2 \).Substituting in the Quotient Rule, we have:\[h'(s) = \frac{(0)(4s^3) - (3)(12s^2)}{(4s^3)^2} = \frac{-36s^2}{16s^6} = \frac{-9}{4s^4}\].

Algebraically manipulate and differentiate

Rewrite the original function \( h(s) = \frac{3}{4s^3} \) as \( h(s) = \frac{3}{4} \cdot s^{-3} \).Differentiate using the power rule, where \( \frac{d}{ds}[s^n] = ns^{n-1} \):\[h'(s) = \frac{3}{4} \cdot (-3) \cdot s^{-3-1} = \frac{-9}{4}s^{-4}\].

Verify equivalence of results from (a) and (b)

From step 1, we obtained \( h'(s) = \frac{-9}{4s^4} \).From step 2, we obtained \( h'(s) = \frac{-9}{4}s^{-4} \), which can also be written as \( \frac{-9}{4s^4} \).Since both forms can be expressed as \( \frac{-9}{4s^4} \), the results from (a) and (b) are equivalent.

Key concepts

Quotient Rule

The Quotient Rule is a technique used in calculus to find the derivative of a function that is the ratio of two differentiable functions. It is particularly useful when dealing with functions in the form of a fraction where one function is divided by another. When applying the Quotient Rule, we define the function in the form \( h(s) = \frac{u(s)}{v(s)} \) where \( u(s) \) and \( v(s) \) are differentiable functions. The derivative of this function, denoted as \( h'(s) \), is given by the formula:

• \[ h'(s) = \frac{u'(s)v(s) - u(s)v'(s)}{[v(s)]^2} \]

To break it down:

• \( u'(s) \) is the derivative of the numerator function \( u(s) \).
• \( v(s) \) is the denominator function.
• \( v'(s) \) is the derivative of the denominator function \( v(s) \).
• \([v(s)]^2\) means the square of the denominator function.

For instance, in our original problem where \( h(s) = \frac{3}{4s^3} \), we set \( u(s) = 3 \) and \( v(s) = 4s^3 \). The derivative \( u'(s) \) becomes 0 because it’s a constant. The calculation with these values then simplifies into the final derivative \( h'(s) = \frac{-9}{4s^4} \). When mastered, the Quotient Rule is a powerful tool that greatly simplifies the differentiation of complex functions.

Power Rule

The Power Rule offers a straightforward method to differentiate terms involving powers of \( s \). It is one of the most fundamental rules in calculus and is used to find the derivative of functions of the form \( s^n \). The rule is expressed as follows:

• \( \frac{d}{ds}[s^n] = n \cdot s^{n-1} \)

This means to differentiate any power of \( s \), you:

• Multiply the power by the coefficient.
• Reduce the power by one.

In our example, the expression \( h(s) = \frac{3}{4} \cdot s^{-3} \) is differentiated using the Power Rule. Performing the differentiation gives:

• \( h'(s) = \frac{3}{4} \times (-3) \times s^{-3-1} = \frac{-9}{4} \times s^{-4} \)

This results in the same expression for the derivative as derived using the Quotient Rule. The Power Rule is beloved for its simplicity and direct application, making it an excellent check for derivative calculations that can otherwise involve more complicated rules.

Algebraic Manipulation

Algebraic manipulation involves rearranging and rewriting expressions to make differentiation easier. It often simplifies the problem at hand, allowing us to avoid more complex rules like the Quotient Rule when unnecessary.
For example, transform \( h(s) = \frac{3}{4s^3} \) into a form more suitable for using the Power Rule. Rewriting it as \( h(s) = \frac{3}{4} \cdot s^{-3} \) achieves this transformation, converting the quotient into a single term with a negative exponent.
Such manipulation is not just a trick for convenience; it ensures mathematical operations remain valid and ultimately lead to the same result. Thus, it is crucial in calculus to optimize the process of finding derivatives efficiently and effectively. By reducing complex expressions, the processes of differentiation becomes simpler, allowing students and practitioners alike to focus on overarching strategies rather than getting bogged down in tedious calculations.