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

Step-by-Step Solution

Answer

The derivative \( h'(3) \) is \(-34\).

Step 1: Find the Derivative of h(x)

The function is given as \( h(x) = 2x + f(x)g(x) \). We want to find the derivative, \( h'(x) \), using differentiation rules. The derivative of \( h(x) \) is given by:\[ h'(x) = \frac{d}{dx}[2x] + \frac{d}{dx}[f(x)g(x)] \]\( \frac{d}{dx}[2x] = 2 \).Now, apply the product rule to differentiate \( f(x)g(x) \), which states that if \( u(x) = f(x) \) and \( v(x) = g(x) \), then \( \frac{d}{dx}[uv] = u'v + uv' \).Thus, \( \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \).Therefore, the derivative is:\[ h'(x) = 2 + f'(x)g(x) + f(x)g'(x) \]

Step 2: Substitute x=3 to Find h'(3)

We need to find \( h'(3) \) by substituting \( x = 3 \) into the derivative expression we found. From the table, \( f(3) = -2 \), \( f'(3) = 8 \), \( g(3) = -4 \), and \( g'(3) = 2 \).Substitute these values into:\[ h'(3) = 2 + f'(3)g(3) + f(3)g'(3) \]\( f'(3)g(3) = 8 \times (-4) = -32 \)\( f(3)g'(3) = (-2) \times 2 = -4 \)Insert these computed values into the equation:\[ h'(3) = 2 - 32 - 4 \]

Step 3: Simplify the Expression

Now simplify the expression calculated in the previous step:\[ h'(3) = 2 - 32 - 4 = -34 \].

Key Concepts

product rule

The product rule is a fundamental technique in calculus used to find the derivative of the product of two differentiable functions. Imagine you have two functions, say \(f(x)\) and \(g(x)\). You want to differentiate their product. This is where the product rule comes into play.
According to the product rule, the derivative of \( f(x)g(x) \) is found by the formula:

• \( \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \)

This rule says you take the derivative of the first function \(f(x)\), multiply it by the second function \(g(x)\), and add it to the first function \(f(x)\) multiplied by the derivative of the second function \(g(x)\).
In simpler terms, think of one function staying untouched while you differentiate the other, and then switch roles. The product rule is essential because it allows us to tackle more complex functions that appear in pairs, making it a powerful tool in derivative computation.

differentiation rules

Differentiation rules are a set of guidelines in calculus that help us find the derivative of functions in a systematic way. They simplify the process of derivative computation by providing specific shortcuts to follow depending on the type of function. Some key differentiation rules include:

• Power Rule: To differentiate \(x^n\), bring down the power \(n\) and reduce the exponent by one: \(\frac{d}{dx}[x^n] = nx^{n-1}\).
• Constant Rule: The derivative of a constant is zero.
• Sum/Difference Rule: The derivative of a sum/difference is the sum/difference of the derivatives.
• Rule for Constants Times a Function: If you have a constant \(c\) multiplied by a function \(f(x)\), the derivative is \(c\) times the derivative of \(f(x)\).

These rules form the foundation of calculus derivatives and often work in tandem. For instance, the original exercise combined the constant rule and the product rule to find the derivative of \( h(x) = 2x + f(x)g(x) \). By mastering these rules, derivative computation becomes much more straightforward.

derivative computation

Derivative computation involves applying differentiation rules to find the rate of change of a function. This process tells us how a function's output changes as its input changes. Whether it's a single variable function or a more complex composition, it all boils down to understanding how each part of the function behaves.
In the given exercise, the function \( h(x) = 2x + f(x)g(x) \), required computing its derivative by breaking it into parts and using appropriate rules. Here’s how this was done:

• The term \(2x\) differentiated using the constant rule yields 2.
• The product \(f(x)g(x)\) applied the product rule for computation.

These calculations produce a new expression that represents the derivative of the original function \( h(x) \). Successfully finding \( h'(3) \) at \(x=3\) showed the detailed process of plugging into the derived formula, which is a common task in calculus.

differentiable functions

A function is differentiable if it has a derivative at every point in its domain. Differentiable functions are smooth, without any jumps or sharp corners, making them predictable and easy to analyze.For any function to be considered differentiable:

• It must be continuous; if a function has discontinuities, it cannot be differentiated everywhere.
• It must not have sharp turns or cusps at the points of interest, as these would produce undefined derivatives there.

In the context of the table given in the exercise, \(f(x)\) and \(g(x)\) are specified as differentiable functions, meaning their derivatives can be computed directly. This ensures that at any value, such as \(x=3\), both the original functions \(f(x), g(x)\) and their derivatives \(f'(x), g'(x)\) are defined, facilitating the calculations involved in derivative computation effectively.