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Second Order Derivatives

Second-order derivatives provide information about the curvature or concavity of a function, indicating how the rate of change of the slope is itself changing. If \(y = f(x)\) is a function, then the first derivative, \(f'(x)\), represents the slope or rate of change of \(y\) with respect to \(x\). The second derivative, denoted as \(f''(x)\), represents the rate of change of the first derivative and provides insight into the function's concavity.

Notation for Second-Order Derivatives

The second derivative of a function can be expressed in several ways: - \( f''(x) \) - \( \frac{d^2y}{dx^2} \) - \( \frac{d^2}{dx^2} f(x) \)

Interpreting the Second Derivative

  1. Concavity of the Function:
  2. If \( f''(x) > 0 \), the function is concave up at that point (the graph is shaped like a "cup").

  3. If \( f''(x) < 0 \), the function is concave down at that point (the graph is shaped like a "cap").

  4. Inflection Points:

  5. An inflection point occurs where the function changes concavity (from concave up to concave down, or vice versa).
  6. At an inflection point, \( f''(x) \) changes sign.

Steps for Finding the Second Derivative

  1. Find the First Derivative:

  2. Differentiate the original function \( f(x) \) to obtain the first derivative, \( f'(x) \).

  3. Differentiate the First Derivative:

  4. Differentiate \( f'(x) \) to obtain the second derivative, \( f''(x) \).

Example 1: Find the Second Derivative of \( f(x) = x^3 - 3x^2 + 2 \)

  1. First Derivative: [ f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 2) = 3x^2 - 6x ]

  2. Second Derivative: [ f''(x) = \frac{d}{dx}(3x^2 - 6x) = 6x - 6 ]

Example 2: Second-Order Derivatives of Parametric Functions

For parametric equations where \( x = f(t) \) and \( y = g(t) \), the second derivative can be found using: [ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}, ] where: - \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \)

Given \( x = t^2 + 1 \) and \( y = t^3 - 3t \):

  1. Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \): [ \frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2 - 3 ]

  2. Find \( \frac{dy}{dx} \): [ \frac{dy}{dx} = \frac{3t^2 - 3}{2t} ]

  3. Differentiate \( \frac{dy}{dx} \) with respect to \( t \): [ \frac{d}{dt}\left(\frac{3t^2 - 3}{2t}\right) = \frac{6t \cdot 2t - (3t^2 - 3) \cdot 2}{(2t)^2} = \frac{12t^2 - 6t^2 + 6}{4t^2} = \frac{6t^2 + 6}{4t^2} ]

  4. Find \( \frac{d^2y}{dx^2} \): [ \frac{d^2y}{dx^2} = \frac{6t^2 + 6}{4t^2 \cdot 2t} = \frac{6(t^2 + 1)}{8t^3} = \frac{3(t^2 + 1)}{4t^3} ]

Example 3: Find the Second Derivative of \( f(x) = e^x \sin(x) \)

  1. First Derivative: Use the product rule: \( f'(x) = e^x \sin(x) + e^x \cos(x) \)

  2. Second Derivative: Again use the product rule: [ f''(x) = \frac{d}{dx}(e^x \sin(x) + e^x \cos(x)) = e^x \sin(x) + e^x \cos(x) + e^x \cos(x) - e^x \sin(x) = 2e^x \cos(x) ]

Second-order derivatives are valuable in analyzing the shape and behavior of functions, particularly in determining local maxima and minima, as well as points of inflection.

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