91Ó°ÊÓ

Differentiation is a fundamental process in calculus that deals with finding the rate at which a function is changing at any given point. It is essentially the operation of computing a derivative, which is a measure of how a function value changes as the variable changes. The derivative is often represented as \( f'(x) \) for a function \( f(x) \). There are multiple rules to find derivatives, and among the most used ones is the Product Rule, which applies when you are finding the derivative of two functions multiplied together.

Understanding differentiation requires familiarity with functions, limits, and the various differentiation rules like the Power Rule (for powers of \( x \)), the Sum Rule (for sums of functions), and the Product Rule. It’s crucial to master these rules to tackle more advanced calculus problems efficiently.

Expanding polynomials is the process of simplifying an expression so that no parentheses are left. We use the distributive property of multiplication over addition to multiply each term in the first polynomial by each term in the second polynomial. In the context of the exercise, \( f(x) = (x - 1)(3x + 4) \) is a product of two binomials.

When we expand the product, we calculate:\[ x(3x) + x(4) - 1(3x) - 1(4) \], which simplifies to \( 3x^2 + 4x - 3x - 4 \). Combining the like terms results in the expanded form \( 3x^2 + x - 4 \). Polynomials are easier to work with and differentiate once they are in their expanded form, especially with elementary derivatives.

The derivative of a product of two functions is found using the Product Rule. 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 \( u'(x)v(x) + u(x)v'(x) \). In simpler terms, this rule tells us that the derivative of a product is not merely the product of the derivatives.

Applying the Product Rule in our exercise, we take the functions \( u(x)=x-1 \) and \( v(x)=3x+4 \) to compute the derivative as:\[ f'(x) = (1)(3x+4) + (x-1)(3) = 6x + 1 \]. The ability to apply the Product Rule correctly is essential in calculus, as it enables you to find derivatives of more complex functions where simply multiplying out might not be feasible.

Verification of the derivative is the step where you confirm the accuracy of the derivative you've computed, usually by comparing it with another method or result. In our problem, we find the derivative using the Product Rule and by expanding the polynomial and differentiating. We end up with the same result, \( f'(x) = 6x + 1 \) from both methods, which verifies our computations.

The importance of verification lies in the confidence it gives you about the procedure and result. It can also serve as a check against possible errors, especially when dealing with more complex functions or when applying advanced calculus techniques. This step is not just a formality but a crucial part of problem-solving in mathematics to ensure solutions are both accurate and reliable.