Problem 126 - 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 \(h^{\prime}(1)\) if \(h(x)=x f(x)+4 g(x)\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Rule for Differentiation

We need to differentiate the function \(h(x) = x f(x) + 4 g(x)\). To do this, we'll use the product rule for the first term \(x f(x)\), which states that if \(u(x) = x\) and \(v(x) = f(x)\), then \((uv)' = u'v + uv'\). The second term, \(4g(x)\), will be differentiated using the constant multiple rule.

Step 2: Differentiate the First Term Using the Product Rule

The first term is \(x f(x)\). Using the product rule: \(u = x\), so \(u' = 1\), and \(v = f(x)\), so \(v' = f'(x)\). Therefore, \((x f(x))' = (1)f(x) + x f'(x) = f(x) + x f'(x)\).

Step 3: Differentiate the Second Term Using the Constant Rule

The second term is \(4 g(x)\). Differentiating this using the constant multiple rule gives \((4 g(x))' = 4 g'(x)\).

Step 4: Combine the Differentiated Terms

Combine the results from Steps 2 and 3 to get the derivative of \(h(x)\): \[h'(x) = f(x) + x f'(x) + 4 g'(x)\].

Step 5: Substitute x = 1 into the Derivative

We need to find \(h'(1)\). Substitute \(x = 1\) into the expression for \(h'(x)\): \[h'(1) = f(1) + 1 \cdot f'(1) + 4 \cdot g'(1)\].

Step 6: Use Values from the Table

From the table, \(f(1) = 3\), \(f'(1) = -1\), and \(g'(1) = 4\). Substitute these values into the expression from Step 5:\[h'(1) = 3 + 1 \cdot (-1) + 4 \cdot 4 = 3 + (-1) + 16\].

Step 7: Calculate the Final Expression

Solve \[3 + (-1) + 16\]. Simplify the expression:\[h'(1) = 3 - 1 + 16 = 18\].

Key Concepts

Product Rule

When working with functions involving products, one of the key differentiation techniques is the product rule. The product rule is essential when you have two functions multiplied together, and you want to find their derivative. In the case of the function \( h(x) = x f(x) + 4 g(x) \), the product rule is applied to the term \( x f(x) \).The product rule states that if you have two functions \( u(x) \) and \( v(x) \), their derivative is given by:\[(uv)' = u'v + uv'\]Here's how it applies:

• Let \( u(x) = x \) and \( v(x) = f(x) \). Then \( u'(x) = 1 \) because the derivative of \( x \) with respect to \( x \) is 1.


• For \( v(x) = f(x) \), we have \( v'(x) = f'(x) \).


• Using the product rule, \( (x f(x))' = u'(x) v(x) + u(x) v'(x) = f(x) + x f'(x) \).

This method systematically allows you to find the derivative when two differentiable functions are multiplied together.

Differentiable Functions

Understanding differentiable functions is crucial in calculus. A function is said to be differentiable at a point if it has a defined derivative at that point. Differentiable functions are smooth and have no sharp corners or discontinuities. For our exercise, we assume that \( f(x) \) and \( g(x) \) are both differentiable, meaning they have defined derivatives denoted by \( f'(x) \) and \( g'(x) \), respectively. These derivatives can be found in the table, which provides their specific values at given points.Differentiability allows us to apply rules like the product rule and the constant multiple rule with confidence, knowing that the functions behave nicely enough for these operations. If a function is not differentiable, these rules cannot be applied directly.Recognizing differentiability is key in calculus to perform various operations and further analyze functions.

Constant Multiple Rule

The constant multiple rule in differentiation is a straightforward yet powerful tool. This rule states that if you have a constant multiplied by a function, the derivative of this product is simply the constant multiplied by the derivative of the function.Consider the term \( 4 g(x) \) from our function \( h(x) = x f(x) + 4 g(x) \). The constant multiple rule tells us:\[(4 g(x))' = 4 g'(x)\]Here's the breakdown:

• The derivative of a constant \( c \) times a function, \( c g(x) \), is \( c g'(x) \).


• In our example, \( c = 4 \) and the differentiated result is \( 4 g'(x) \).

This rule allows easy differentiation when constants are involved, ensuring that only the function part is differentiated. It simplifies the process of finding derivatives and is often the first step in more complex differentiation problems.