Calculating Second Derivatives A Step By Step Guide
This article delves into the process of finding the second derivative, denoted as f''(x), for two distinct functions. Understanding the second derivative is crucial in calculus as it provides insights into the concavity and inflection points of a function. We will explore the step-by-step calculations for each function, ensuring clarity and comprehension.
1. Finding the Second Derivative of f(x) = √(x² + 3x)
To determine the second derivative of the function f(x) = √(x² + 3x), we will follow a systematic approach involving the chain rule and quotient rule of differentiation. The second derivative, f''(x), represents the rate of change of the first derivative, f'(x), and provides valuable information about the function's concavity. In this section, we will meticulously calculate f''(x) for the given function, elucidating each step for clarity. Our main objective here is to find out how the rate of change of the slope of the tangent line to the curve f(x) is changing. This involves a detailed application of calculus principles.
Step 1: Find the First Derivative f'(x)
We begin by finding the first derivative, f'(x), of the function f(x) = √(x² + 3x). This requires the application of the chain rule. Recall that the chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In this case, the outer function is the square root function, and the inner function is x² + 3x. So, we can rewrite f(x) as (x² + 3x)^(1/2). Applying the power rule and the chain rule, we get:
f'(x) = (1/2)(x² + 3x)^(-1/2) * (2x + 3)
Simplifying this expression, we obtain:
f'(x) = (2x + 3) / (2√(x² + 3x))
This is our first derivative. It represents the slope of the tangent line to the curve of f(x) at any given point. Now that we have f'(x), we can proceed to find the second derivative.
Step 2: Find the Second Derivative f''(x)
To find the second derivative, f''(x), we need to differentiate f'(x) = (2x + 3) / (2√(x² + 3x)). This requires the quotient rule. The quotient rule states that the derivative of a quotient of two functions is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. Applying the quotient rule, we have:
f''(x) = [2√(x² + 3x) * (2) - (2x + 3) * (2 * (1/2)(x² + 3x)^(-1/2) * (2x + 3))] / [4(x² + 3x)]
Now, let's simplify this expression step by step. First, we can simplify the term inside the brackets:
f''(x) = [4√(x² + 3x) - (2x + 3)² / √(x² + 3x)] / [4(x² + 3x)]
To further simplify, we can multiply the numerator and denominator by √(x² + 3x) to eliminate the fraction within the numerator:
f''(x) = [4(x² + 3x) - (2x + 3)²] / [4(x² + 3x)^(3/2)]
Expanding the terms in the numerator, we get:
f''(x) = [4x² + 12x - (4x² + 12x + 9)] / [4(x² + 3x)^(3/2)]
Simplifying the numerator, we have:
f''(x) = -9 / [4(x² + 3x)^(3/2)]
This is the second derivative of the function f(x) = √(x² + 3x). The negative sign indicates that the function is concave down in the interval where it is defined.
Step 3: Interpretation of f''(x)
The second derivative, f''(x) = -9 / [4(x² + 3x)^(3/2)], provides valuable information about the concavity of the function f(x) = √(x² + 3x). Since f''(x) is negative for all x in the domain of f(x) (where x² + 3x > 0), the function is concave down in its domain. Concavity is a measure of how the curve of a function bends. A concave down function curves downward, while a concave up function curves upward. In this case, the negative second derivative indicates that the graph of f(x) is always bending downwards.
Furthermore, inflection points occur where the second derivative changes sign. However, since f''(x) is always negative in the domain of f(x), there are no inflection points for this function. Inflection points are points where the concavity of a function changes. Understanding the concavity and inflection points helps us to sketch the graph of the function and analyze its behavior.
2. Finding the Second Derivative of f(x) = x / (x² + 4)
Now, let's find the second derivative of the function f(x) = x / (x² + 4). This function is a rational function, and finding its derivatives will involve the quotient rule. The quotient rule is a fundamental tool in calculus for differentiating functions that are expressed as a ratio of two other functions. As before, we will proceed in a step-by-step manner, first finding the first derivative and then differentiating it again to obtain the second derivative.
Step 1: Find the First Derivative f'(x)
To find the first derivative f'(x) of f(x) = x / (x² + 4), we apply the quotient rule. Let u(x) = x and v(x) = x² + 4. Then, u'(x) = 1 and v'(x) = 2x. Applying the quotient rule, we have:
f'(x) = [v(x)u'(x) - u(x)v'(x)] / [v(x)]²
Substituting the expressions for u(x), v(x), u'(x), and v'(x), we get:
f'(x) = [(x² + 4)(1) - x(2x)] / (x² + 4)²
Simplifying the numerator, we have:
f'(x) = (x² + 4 - 2x²) / (x² + 4)²
f'(x) = (4 - x²) / (x² + 4)²
This is the first derivative of the function. It tells us about the slope of the function at any point x.
Step 2: Find the Second Derivative f''(x)
To find the second derivative, f''(x), we need to differentiate f'(x) = (4 - x²) / (x² + 4)². This again requires the quotient rule, but this time the expressions are more complex. Let u(x) = 4 - x² and v(x) = (x² + 4)². Then, u'(x) = -2x. To find v'(x), we use the chain rule: v'(x) = 2(x² + 4)(2x) = 4x(x² + 4). Applying the quotient rule, we have:
f''(x) = [v(x)u'(x) - u(x)v'(x)] / [v(x)]²
Substituting the expressions for u(x), v(x), u'(x), and v'(x), we get:
f''(x) = [(x² + 4)²(-2x) - (4 - x²)(4x(x² + 4))] / [(x² + 4)²]²
f''(x) = [-2x(x² + 4)² - 4x(4 - x²)(x² + 4)] / (x² + 4)⁴
We can factor out a common factor of -2x(x² + 4) from the numerator:
f''(x) = [-2x(x² + 4)((x² + 4) + 2(4 - x²))] / (x² + 4)⁴
Simplifying the expression inside the parentheses, we get:
f''(x) = [-2x(x² + 4)(x² + 4 + 8 - 2x²)] / (x² + 4)⁴
f''(x) = [-2x(x² + 4)(12 - x²)] / (x² + 4)⁴
We can cancel out one factor of (x² + 4) from the numerator and denominator:
f''(x) = [-2x(12 - x²)] / (x² + 4)³
f''(x) = [2x(x² - 12)] / (x² + 4)³
This is the second derivative of the function f(x) = x / (x² + 4).
Step 3: Interpretation of f''(x)
The second derivative, f''(x) = [2x(x² - 12)] / (x² + 4)³, provides information about the concavity of the function f(x) = x / (x² + 4). By analyzing the sign of f''(x), we can determine the intervals where the function is concave up or concave down.
To find the inflection points, we need to find the values of x where f''(x) = 0 or is undefined. f''(x) = 0 when 2x(x² - 12) = 0, which occurs at x = 0, x = √12 = 2√3, and x = -√12 = -2√3. The denominator (x² + 4)³ is never zero, so f''(x) is defined for all real numbers.
Now, we analyze the sign of f''(x) in the intervals determined by these critical points:
- For x < -2√3, f''(x) < 0, so the function is concave down.
- For -2√3 < x < 0, f''(x) > 0, so the function is concave up.
- For 0 < x < 2√3, f''(x) < 0, so the function is concave down.
- For x > 2√3, f''(x) > 0, so the function is concave up.
Therefore, the inflection points occur at x = -2√3, x = 0, and x = 2√3. These points mark the transitions between concave up and concave down sections of the graph of f(x). Understanding the second derivative and its implications allows us to sketch the graph of the function with greater accuracy and to analyze its behavior in detail.
Conclusion
In this article, we have successfully found the second derivatives of two functions: f(x) = √(x² + 3x) and f(x) = x / (x² + 4). For the first function, we found that f''(x) = -9 / [4(x² + 3x)^(3/2)], which indicates that the function is concave down in its domain. For the second function, we found that f''(x) = [2x(x² - 12)] / (x² + 4)³, and we identified the inflection points at x = -2√3, x = 0, and x = 2√3. The process of finding the second derivative involves applying the rules of differentiation, including the chain rule and the quotient rule, and simplifying the resulting expressions. The second derivative provides valuable information about the concavity and inflection points of a function, which are crucial for understanding its behavior and sketching its graph. This detailed analysis demonstrates the power of calculus in exploring the properties of functions.