Problem 127 - Calculus Volume 1

Question

Assume that \(f(x)\) and \(g(x)\) are both differentiable functions with values as given in the following table. Use the following table to calculate the following derivatives. $$ \begin{array}{|c|c|c|c|c|} \hline \boldsymbol{x} & 1 & 2 & 3 & 4 \\ \hline \boldsymbol{f}(\boldsymbol{x}) & 3 & 5 & -2 & 0 \\ \hline \boldsymbol{g}(\boldsymbol{x}) & 2 & 3 & -4 & 6 \\ \hline \boldsymbol{f}^{\prime}(\boldsymbol{x}) & -1 & 7 & 8 & -3 \\ \hline \boldsymbol{g}^{\prime}(\boldsymbol{x}) & 4 & 1 & 2 & 9 \\ \hline \end{array} $$ Find \(h^{\prime}(2)\) if \(h(x)=\frac{f(x)}{g(x)}\).

Step-by-Step Solution

Answer

The derivative \( h'(2) \) is \( \frac{16}{9} \).

Step 1: Find the appropriate formula

To find the derivative of a quotient of two functions, we use the quotient rule. The formula is: \[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]This rule applies because the function \( h(x) \) given is defined as a quotient, \( h(x) = \frac{f(x)}{g(x)} \).

Step 2: Substitute known values into the formula

We need to find \( h'(2) \). Using the formula from Step 1, substitute the known values:- \( f(2) = 5 \)- \( g(2) = 3 \)- \( f'(2) = 7 \)- \( g'(2) = 1 \)Plug these into the formula:\[ h'(2) = \frac{7 \cdot 3 - 5 \cdot 1}{3^2} \].

Step 3: Calculate the numerator and denominator

Calculate the numerator of the derivative:\[ 7 \cdot 3 - 5 \cdot 1 = 21 - 5 = 16 \]Calculate the denominator:\[ 3^2 = 9 \]

Step 4: Compute the derivative

Now that we have both parts of the fraction, compute the value of \( h'(2) \):\[ h'(2) = \frac{16}{9} \].

Key Concepts

Understanding Derivatives

Derivatives are fundamental in the world of calculus. They measure how a function changes as its input changes. Think of it as a way to determine the "instantaneous rate of change" or the slope of the function at any given point. In simpler terms, if you were riding a bike up a hill, the derivative would tell you how steep the hill is at each point.
For functions like \( f(x) \) and \( g(x) \), their derivatives \( f'(x) \) and \( g'(x) \) are the key to understanding the behavior of these functions.

• The derivative \( f'(x) \) tells us how \( f(x) \) is changing.
• Similarly, \( g'(x) \) shows the rate at which \( g(x) \) changes.

By using derivatives, we can solve various problems related to rates of change, optimization, and more complex dynamical systems. They are useful in physics, engineering, economics, and beyond.

Exploring Differentiable Functions

A function being differentiable means it has a derivative at every point in its domain. In other words, the function is smooth; it doesn't have any sharp corners or jumps that would make taking a derivative impossible.
This smoothness implies two things:

• The function is continuous, meaning you could draw it without lifting your pen.
• The function has no abrupt changes in direction.

When you have differentiable functions like \( f(x) \) and \( g(x) \), you know you can confidently apply calculus techniques to analyze them. The quotient rule, which we used in the solution, is a perfect example of using the differentiability of two functions to find the rate of change of their quotient. Having \( f(x) \) and \( g(x) \) differentiable assures us that their quotient \( h(x) = \frac{f(x)}{g(x)} \) is differentiable as well, provided that \( g(x) eq 0 \). This allows us to calculate \( h'(x) \) using the smooth functions \( f(x) \) and \( g(x) \) easily.

Connecting Concepts in Calculus

Calculus is the branch of mathematics that deals with continuous change. It is divided mainly into two parts: differential calculus and integral calculus. In the context of this exercise, we are focusing on differential calculus, which involves finding derivatives.
Differential calculus is crucial because it helps us get insights into the geometry of curves, optimization problems, and real-world dynamics.
Within this realm, we come across several important rules like the quotient rule, which aids in differentiating functions presented as fractions. The quotient rule formula is:\[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]This rule is handy when you need to find the derivative of a quotient of two differentiable functions at a specific point, just like in our example.
Ultimately, calculus provides tools not only to analyze and model changes but also to solve complex problems in science and engineering. It is a powerful discipline that extends its concepts to solving practical problems in diverse fields.