Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$ f(x)=\left(x^{2}+1\right)(2 x+5) \quad(-1,6) $$

Short answer

Hence, the value of the derivative of the function at the point (-1,6) is -2, and the differentiation rule used to find the derivative is the Product Rule.

Step-by-step solution

Identify the two functions

Here, the two functions are \(u(x)=x^{2}+1\) and \(v(x)=2x+5\).

Find the derivatives of the two functions

The derivative of \(u(x)\), denoted as \(u'(x)\), is \(2x\). The derivative of \(v(x)\), denoted as \(v'(x)\), is \(2\). So we have \(u'(x)=2x\) and \(v'(x)=2\).

Apply the Product Rule

Applying the Product rule to find the derivative \(f'(x)\) of the function, we have: \(f'(x) = u(x)v'(x) + v(x)u'(x) = (x^{2}+1)(2) + (2x+5)(2x) = 2x^{2} +2 + 4x^{2} +10x = 6x^{2} +10x +2\).

Substitute the given x-value into the derivative

Substitute x = -1 into \(f'(x)\) to get \(f'(-1) = 6(-1)^{2} +10(-1) +2 = 6-10+2 = -2\).

Key concepts

Derivative of a Function

Understanding the concept of a derivative is fundamental to calculus. The derivative represents the rate at which a function is changing at any given point and is a key player in diverse fields, from physics to economics. For a function f(x), the derivative at a specific point is the slope of the tangent line to the function's graph at that point. Think of it as how fast the y-value of the function is changing per unit of the x-value at that particular spot.

When we talk about the derivative of a function, we're often interested in finding a formula for the derivative, which we denote with f'(x), that defines this rate of change for every x-value in the function's domain. In our exercise, we initially have f(x) = (x^2 + 1)(2x + 5), and our task is to find f'(x). This leads us naturally to the application of differentiation rules to manage the product of two functions systematically.

Applying Differentiation Rules

Differentiation rules, also known as differentiation techniques, are the established methods or shortcuts that simplify the process of finding derivatives. One such rule is the Product Rule, which is vital when dealing with the multiplication of two functions, as evidenced in our exercise.

The Product Rule states that if you have two differentiable functions u(x) and v(x), the derivative of their product u(x)v(x) is given by u'(x)v(x) + u(x)v'(x). Here's a handy mnemonic: the derivative of the product is the derivative of the first times the second, plus the first times the derivative of the second.

In the provided solution, u(x) is x^2 + 1, and v(x) is 2x + 5. The Product Rule application gives us the derivative f'(x) = (x^2+1)(2) + (2x+5)(2x), which simplifies to 6x^2 +10x +2. This step of applying the Product Rule is what ensures an accurate and swift derivation of the product of functions.

Calculus Problem Solving

Calculus problem solving requires a methodical approach to dissect complex problems into manageable parts. Initially, identifying the functions involved and their derivatives sets a solid foundation. Consider this a routine 'setup phase' for the problem-solving journey.

With the foundations laid, the next logical step is the application of relevant differentiation rules, as seen with the Product Rule in our exercise. Efficient calculus problem solving involves recognizing which rule to apply and when. Meticulous application of these rules, followed by simplifying the resulting expressions, leads to a derivative function that holds true for all values in its domain.

The final act in our calculus theater is plugging the x-value from the given point into the derived function to find the actual derivative value at that point. The connection between abstract differentiation and concrete numbers marks the conclusion of the problem-solving process. In our exercise, substituting x = -1 renders the derivative at that point as f'(-1) = -2. Not only has the derivative been found, but a particular instance of its application has been realized, showcasing the full scope of calculus problem solving.