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Mathematical Logic
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Differentiation
Applications of Derivatives
Integration
Definite Integration
Applications of Definite Integration
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Differential Equation and Applications
- Basic Concepts of Differential Equations
- Order and Degree of a Differential Equation
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- Methods of Solving Differential Equations> Variable Separable Differential Equations
- Methods of Solving Differential Equations> Homogeneous Differential Equations
- Methods of Solving Differential Equations>Linear Differential Equations
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Commission, Brokerage and Discount
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Insurance and Annuity
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Linear Regression
- Regression
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- Lines of Regression of X on Y and Y on X Or Equation of Line of Regression
- Properties of Regression Coefficients
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Time Series
- Introduction to Time Series
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Index Numbers
- Weighted Aggregate Method
- Cost of Living Index Number
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Linear Programming
Assignment Problem and Sequencing
- Assignment Problem
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Probability Distributions
Definition: Second Order Derivative
Let y = f(x). If the first derivative \[\frac{dy}{dx} = f'(x)\] is itself differentiable, then differentiating once again with respect to \[x\] gives the second order derivative.
Example 1
If \[y = \text{A} \sin x + \text{B} \cos x\], then prove that \[\frac{d^{2}y}{dx^{2}} + y = 0\].
Solution: We have
and
Hence \[\frac{d^{2}y}{dx^{2}} + y = 0\]
Example 2
If \[y = \sin^{-1} x\], show that \[(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0\].
Solution: We have \[y = \sin^{-1} x\]. Then
or
So
\[\frac{d}{dx} \left( \sqrt{(1 - x^2)} \cdot \frac{dy}{dx} \right) = 0\]
or
or
Hence \[(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0\]
Alternatively, Given that \[y = \sin^{-1} x\], we have
So\[(1 - x^2) \cdot 2y_1 y_2 + y_1^2 (0 - 2x) = 0\]
Hence \[(1 - x^2) y_2 - xy_1 = 0\]
Key Points: Second Order Derivative
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Second derivative means differentiating the function twice with respect to the same variable.
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It is defined only when the first derivative is differentiable.
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Common notations are \[\frac{d^2y}{dx^2}\], f''(x), y'', \[D^2y\], and \[y_2\].
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Higher order derivatives can be defined similarly.
