Problem 126 - Calculus Volume 1

Question

For the following exercises, 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. $$\begin{array}{|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} \\ \hline f(x) & {3} & {5} & {-2} & {0} \\ \hline g(x) & {2} & {3} & {-4} & {6} \\ \hline f^{\prime}(x) & {-1} & {7} & {8} & {-3} \\ \hline g^{\prime}(x) & {4} & {1} & {2} & {9} \\ \hline\end{array}$$ Find \(h^{\prime}(1)\) if \(h(x)=x f(x)+4 g(x)\)

Step-by-Step Solution

Answer

The derivative \( h'(1) = 18 \).

Step 1: Substitute Values into the Function

The function given is \( h(x) = x f(x) + 4 g(x) \). We first need to differentiate this function before substituting values. So let's perform the differentiation.

Step 2: Differentiate the Function

The derivative of \( h(x) = x f(x) + 4 g(x) \) can be found using the product rule and sum rule. - For \( x f(x) \), use the product rule: \((uv)' = u'v + uv'\).- For \( 4g(x) \), use the constant multiple rule: \((cu)' = cu'\). Thus, the derivative of \( h(x) \) is: \[ h'(x) = (x f(x))' + (4 g(x))' = (1)f(x) + x f'(x) + 4 g'(x) \].

Step 3: Substitute Specific Values into the Derivative

Now, substitute \( x = 1 \) into the derivative \( h'(x) = f(x) + x f'(x) + 4 g'(x) \). Using the information from the table: - \( f(1) = 3 \)- \( f'(1) = -1 \)- \( g'(1) = 4 \) Thus, we have: \[ h'(1) = 3 + (1)(-1) + 4(4) \].

Step 4: Calculate the Result

Compute the expression: - First, calculate \( 1 \times -1 = -1 \). - Then, calculate \( 4 \times 4 = 16 \). - Finally, sum them up: \[ h'(1) = 3 - 1 + 16 = 18 \].

Key Concepts

Differentiable Functions

Differentiable functions are central to calculus. A function is said to be differentiable at a point if it has a derivative there. This implies that the function should be smooth and continuous at that point, with no sharp turns or breaks. Differentiation, which refers to finding a derivative, allows us to understand how a function changes over its domain.
In practical terms, if you can draw a function with a pencil without lifting it off the paper, it’s likely differentiable at that point. Differentiability ensures that we can apply calculus rules to analyze such functions. Understanding this concept is essential, as it forms the foundation for using various rules of differentiation.

Product Rule

The product rule is a crucial tool for finding derivatives when dealing with products of two differentiable functions. If you have two functions, say \(u(x)\) and \(v(x)\), the derivative of their product is not just the product of their derivatives. Instead, it is given by:

• \((uv)' = u'v + uv'\)

This means you take the derivative of the first function, multiply it by the second, then add the first function times the derivative of the second.
In the given exercise, the product rule was applied to \(x f(x)\). By applying the product rule, we help account for the changes in both components of the product as they affect the overall function.

Sum Rule

The sum rule simplifies the differentiation of sums of functions. It states that the derivative of the sum of two functions is the sum of their derivatives. In formula terms, if you have two functions \(u(x)\) and \(v(x)\), then:

• \((u + v)' = u' + v'\)

This principle is straightforward but powerful, allowing us to break down complex expressions into manageable parts. The derivative of the sum of functions is just as straightforward as adding their individual derivatives. In the problem you provided, this rule was part of differentiating \(h(x) = x f(x) + 4 g(x)\), allowing us to treat each part independently before summing back.

Constant Multiple Rule

The constant multiple rule is a basic yet important concept in differentiation. This rule states that when you multiply a function by a constant, you can keep the constant and only differentiate the function. Mathematically, if \(c\) is a constant and \(u(x)\) is a function, then:

• \((cu)' = cu'\)

This rule simplifies many differentiation tasks, as it allows constants to remain "outside" the differentiation process, keeping focus only on the function.
For example, in the exercise where \(h(x) = 4 g(x)\), the derivative was simply \(4 g'(x)\), easily found by keeping the constant 4 intact while differentiating \(g(x)\). This rule is vital for efficiently working through derivatives of more complicated functions.