Question

Use the given information to find \(f^{\prime}(2)\). \(g(2)=3 \quad\) and \(\quad g^{\prime}(2)=-2\) \(h(2)=-1 \quad\) and \(\quad h^{\prime}(2)=4\) $$ f(x)=g(x) h(x) $$

Short answer

\The value of \(f^{\prime}(2)\) is 14.

Step-by-step solution

Identify the given values

The values given in the exercise are: \(g(2) = 3\), \(g'(2) = -2\), \(h(2) = -1\), and \(h'(2) = 4\). The function \(f(x)\) is the product of \(g(x)\) and \(h(x)\).

Apply the Product Rule

The product rule for differentiation is: \((f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\). Since \(f(x) = g(x)h(x)\), we can write the derivative \(f'(x)\) as: \(f'(x) = g'(x)h(x) + g(x)h'(x)\).

Substitute the given values into the derivative expression

Now we can substitute \(x = 2\) into \(f'(x) = g'(x)h(x) + g(x)h'(x)\) to get \(f'(2) = g'(2)h(2) + g(2)h'(2)\). Then, by substituting the given function values we get \(f'(2) = (-2)(-1) + (3)(4) = 2 + 12\).

Key concepts

Calculus

Calculus is the branch of mathematics that deals with rates of change and accumulation. At its core, calculus is built upon two fundamental concepts: differentiation and integration. Differentiation focuses on determining how a function changes at any given point, and it provides a systematic way to compute the rate at which quantities change. This is crucial in a wide range of applications, from motion in physics to optimization in economics. Calculus is a powerful tool because it helps us make sense of complex systems by breaking them down into relationships between variables that are changing over time.

When dealing with functions, calculus allows us to find the slope of the function at any point, which is what we call the derivative. For example, if you have a function that describes the distance a car travels over time, the derivative of this function would give you the car's speed at any given moment. This concept is what lies at the heart of problems such as the one we're exploring, where we want to find the derivative of a product of two functions at a specific point.

Derivative Calculation

The process of calculating derivatives, known as derivative calculation, is a fundamental operation in calculus. It involves using differentiation rules and formulas to determine the rate of change of a function. There are several techniques to calculate derivatives, such as using the definition of the derivative as the limit of the difference quotient, applying rules like the power rule, product rule, and chain rule, and using higher-order derivatives for more complex functions.

Example of Derivative Calculation

In the given exercise, we are asked to find the derivative of the function f(x) = g(x)h(x) at the point x = 2. To handle this task, we first identify known values for g(2), g'(2), h(2), and h'(2), and then apply the product rule, a specific technique in derivative calculation, to find our answer.

Chain Rule

The chain rule is a crucial differentiation technique in calculus that permits us to find the derivative of a composite function. In simple terms, if a function y depends on a variable u, which itself depends on another variable x, the chain rule helps us to find dy/dx, even though we need to go through u. This is done by multiplying the derivative of y with respect to u by the derivative of u with respect to x, or (dy/du) * (du/dx).

The chain rule is essential when dealing with functions that have multiple layers of functions within them, like f(g(x)). It essentially allows us to 'chain' together the derivatives of the outer and inner functions. To apply the chain rule, one must be able to identify the inner function(s) and differentiate each function separately before combining them.

Differentiation Techniques

There are several differentiation techniques that are used to tackle various types of functions. For instance, the product rule, as used in the exercise, is a specific method to differentiate functions that are multiplied together. Other techniques include the quotient rule for division, the power rule for powers of x, and the sum rule for the sum of functions.

Importance of Differentiation Techniques

Understanding these techniques is central to calculus, as they are applied to solve a myriad of problems involving rates of change. Each rule has its own specific use-case, and mastering when and how to apply each rule is crucial to tackling calculus problems effectively. For the product rule, its primary use is when differentiating the product of two or more functions, which states that d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x). With practice, students learn to identify which rule to apply in different scenarios, streamlining the process of derivative calculation.