Problem 248 - Calculus Volume 1

Question

Use the information in the following table to find \(h^{\prime}(a)\) at the given value for \(a\). $$\begin{array}{|c|c|c|c|c|} \hline x & f(x) & f^{\prime}(x) & g(x) & g^{\prime}(x) \\ \hline 0 & 2 & 5 & 0 & 2 \\ \hline 1 & 1 & -2 & 3 & 0 \\ \hline 2 & 4 & 4 & 1 & -1 \\ \hline 3 & 3 & -3 & 2 & 3 \\ \hline \end{array}$$ $$ h(x)=\left(\frac{f(x)}{g(x)}\right)^{2} ; a=3 $$

Step-by-Step Solution

Answer

\( h'(3) = \frac{-45}{4} \).

Step 1: Determine the Formula for Derivative

For the given function \( h(x) = \left(\frac{f(x)}{g(x)}\right)^2 \), we need to find its derivative \( h'(x) \). To do this, we will apply the chain rule and the quotient rule. Let \( u(x) = \frac{f(x)}{g(x)} \). Then \( h(x) = u(x)^2 \). The derivative \( h'(x) \) is given by \( 2u(x) \cdot u'(x) \).

Step 2: Apply the Quotient Rule

To find \( u'(x) \), we apply the quotient rule: \( u'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \). This gives us the derivative of \( u(x) = \frac{f(x)}{g(x)} \).

Step 3: Substitute Given Values (Find \( u(3) \))

First, calculate \( u(3) \) using the values from the table: \( u(3) = \frac{f(3)}{g(3)} = \frac{3}{2} \).

Step 4: Calculate \( u'(3) \)

Use the quotient rule with \( x = 3 \): \( f(3) = 3 \), \( g(3) = 2 \), \( f'(3) = -3 \), and \( g'(3) = 3 \). Thus, \( u'(3) = \frac{(-3)(2) - (3)(3)}{2^2} = \frac{-6 - 9}{4} = \frac{-15}{4} \).

Step 5: Calculate \( h'(3) \)

Now substitute the values into the chain rule formula for \( h'(x) \): \( h'(3) = 2u(3) \cdot u'(3) = 2 \cdot \frac{3}{2} \cdot \left(\frac{-15}{4}\right) = 3 \cdot \frac{-15}{4} = \frac{-45}{4} \).

Step 6: Final Result

The derivative of \( h(x) \) at \( x = 3 \) is \( h'(3) = \frac{-45}{4} \).

Key Concepts

Quotient Rule

When dealing with derivatives of divisions, the quotient rule becomes an essential tool. Imagine you need to derive a function that is simply the quotient of two other functions, say \(u(x) = \frac{f(x)}{g(x)}\). The quotient rule gives us a systematic way to calculate \(u'(x)\). Here’s what it looks like:

• Start by identifying the numerator function \(f(x)\) and the denominator function \(g(x)\).
• The quotient rule states: \(u'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\).
• Remember, the order of terms matters: \(f'(x)g(x)\) comes first, followed by \(f(x)g'(x)\), subtracted from the former.

In our example, thanks to the table given, we used this rule to find that \(u'(3) = \frac{-15}{4}\). Make sure the denominator function \(g(x)\) is never zero, as division by zero is undefined.

Chain Rule

The chain rule is your friend whenever you deal with compositions of functions. If a function is inside another function and you need its derivative, the chain rule steps in. It might sound tricky, but with a little practice, it becomes intuitive.

• The formula is simple: if you have \(h(x) = v(u(x))\), the derivative is \(h'(x) = v'(u(x)) \cdot u'(x)\).
• Think of it as peeling away each function layer by layer, like an onion, finding the derivative of each layer and multiplying them.

For \(h(x) = \left(\frac{f(x)}{g(x)}\right)^2\), we first took the derivative of the outer function, \(u(x)^2\), which is \(2u(x) \cdot u'(x)\). Then, we calculated \(u'(x)\) separately using the quotient rule. This is how we found \(h'(3) = \frac{-45}{4}\).

Function Derivative

Understanding how derivatives work and why they're useful is central in calculus. A derivative, in essence, represents the rate of change. It shows how a function changes at any given moment. In our exercise, several derivatives were used:

• First, the derivative of \(f(x)\) and \(g(x)\), written as \(f'(x)\) and \(g'(x)\).
• Then, these were used to find \(u'(x)\) and finally \(h'(x)\) as part of more complex derivative calculations.

The goal was to see how the function \(h(x)\) behaves near \(x = 3\). We ended up with a derivative value of \(\frac{-45}{4}\), which indicates the slope or rate of change of our function at that point.

Mathematical Problem Solving

Solving mathematical problems requires a step-by-step logical approach. It’s not just about sense-making; it's about systematically applying rules and formulas to achieve a solution. For derivatives, this means:

• Identifying which rules apply: here, both the chain rule and quotient rule.
• Clearly organizing known values and substituting them methodically, as seen in the example where values from the table were substituted step-by-step.
• Breaking the problem into bite-sized chunks—tackle each piece separately, like finding \(u(x)\) first before \(h(x)\).

Always double-check your work at each stage to ensure accuracy, much like how we checked each \(u(x)\) and \(u'(x)\) calculation before compiling them into the final value of \(h'(3)\). Stay organized, steady, and persistent. That's mathematical problem-solving!