Problem 128 - 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 \(h^{\prime}\) (3) if \(h(x)=2 x+f(x) g(x)\).

Step-by-Step Solution

Answer

\( h^{\prime}(3) = -34 \)

Step 1: Differentiate the Function

Find the derivative of \( h(x) = 2x + f(x)g(x) \) using the product rule. The derivative is \( h^{\prime}(x) = 2 + f^{\prime}(x)g(x) + f(x)g^{\prime}(x) \).

Step 2: Substitute the Values

Substitute \( x = 3 \) into the derivative. From the table, \( f(3) = -2 \), \( g(3) = -4 \), \( f^{\prime}(3) = 8 \), and \( g^{\prime}(3) = 2 \). Substitute these values into \( h^{\prime}(x) \) to get: \[ h^{\prime}(3) = 2 + 8(-4) + (-2)(2). \]

Step 3: Simplify the Expression

Calculate the expression found in Step 2: \( h^{\prime}(3) = 2 + 8(-4) + (-2)(2) = 2 - 32 - 4 \).

Step 4: Final Calculation

Simplify \( 2 - 32 - 4 \) to get \( h^{\prime}(3) = -34 \).

Key Concepts

Product Rule

In calculus, the Product Rule is a fundamental tool for differentiating the product of two functions. When you have a function in the form of the product of two differentiable functions, say \( f(x) \) and \( g(x) \), the derivative is not as simple as differentiating each part separately. Instead, the Product Rule gives us the formula:\[ (fg)^{\prime}(x) = f^{\prime}(x)g(x) + f(x)g^{\prime}(x) \]This formula is essential because it considers how both functions are changing with respect to \( x \). If you think about it, the first part \( f^{\prime}(x)g(x) \) tells us how \( g \) is scaled by the change in \( f \), while \( f(x)g^{\prime}(x) \) tells us how \( f \) is scaled by the change in \( g \).
So, to find \( h^{\prime}(x) = 2x + f(x)g(x) \), apply the Product Rule on just \( f(x)g(x) \). This simplifies the differentiation to: \( h^{\prime}(x) = 2 + f^{\prime}(x)g(x) + f(x)g^{\prime}(x) \).
Always remember this key principle when you encounter products of functions in calculus!

Differentiable Functions

A function being differentiable means that it has a derivative at each point within its domain.
This also implies the function is smooth and continuous, without any sharp corners or breaks.

• For functions \( f(x) \) and \( g(x) \) to be differentiable, they must be smooth and their derivatives \( f^{\prime}(x) \) and \( g^{\prime}(x) \) must exist at the points we're interested in.
• This concept is crucial because if a function is not differentiable at a point, the rules for differentiation, like the product and chain rules, do not apply.
• For instance, in our given problem, because \( f \) and \( g \) are differentiable, we are able to confidently use their derivatives and apply the product rule.

The table in the exercise verifies the differentiability by providing \( f^{\prime}(x) \) and \( g^{\prime}(x) \), thus allowing us to proceed with the differentiation calculations accurately.

Substitution in Calculus

Substitution in calculus is a method used to simplify expressions and solve derivatives or integrals by directly replacing variables with specific values.
This is especially useful when you need to evaluate a derivative at a particular point.In the exercise, once the derivative \( h^{\prime}(x) = 2 + f^{\prime}(x)g(x) + f(x)g^{\prime}(x) \) was calculated, the next step involved substitution. To find \( h^{\prime}(3) \), you substitute \( x = 3 \) into the derivative function. Here's how you proceed:

• Identify the relevant values from the given table: \( f(3) = -2 \), \( g(3) = -4 \), \( f^{\prime}(3) = 8 \), and \( g^{\prime}(3) = 2 \).
• Plug these values into the derivative expression: \( h^{\prime}(3) = 2 + 8(-4) + (-2)(2) \).
• Simplify this to find \( h^{\prime}(3) = -34 \).

This method of substitution makes it easy to evaluate derivatives or integrals at specific points without having to re-calculate from scratch. It's efficient and reduces complexity in calculus problems.