Question

Finding a Pattern Consider the function \(f(x)=\sin \beta x\) where \(\beta\) is a constant. (a) Find the first- second-, third-, and fourth-order derivatives of the function. (b) Verify that the function and its second derivative satisfy the equation \(f^{\prime \prime}(x)+\beta^{2} f(x)=0\) (c) Use the results of part (a) to write general rules for the even- and odd- order derivatives \(f^{(2 k)}(x)\) and \(f^{(2 k-1)}(x) .\) \(\left[\text { Hint: }(-1)^{k} \text { is positive if } k \text { is even and negative if } k \text { is }\right.\) odd. \(]\)

Short answer

The first four derivatives of \(f(x) = \sin(\beta x)\) are \(f'(x) = \cos(\beta x) * \beta\), \(f''(x) = -\sin(\beta x) * \beta^{2}\), \(f'''(x) = -\cos(\beta x) * \beta^{3}\), and \(f''''(x) = \sin(\beta x) * \beta^{4}\). The equation \(f''(x) + \beta^{2} * f(x) = 0\) holds when substituted with the function and its second derivative. The general rules for even or odd order derivatives are: for even k, \(f^{(2k)}(x) = \sin(\beta x) * \beta^{2k}\) and \(f^{(2k-1)}(x) = \cos(\beta x) * \beta^{2k-1}\); for odd k, \(f^{(2k)}(x) = -\sin(\beta x) * \beta^{2k}\) and \(f^{(2k-1)}(x) = -\cos(\beta x) * \beta^{2k-1}\).

Step-by-step solution

Find the First four Derivatives

First, let's find the first four derivatives of the function \(f(x) = \sin(\beta x)\)\n\nThe first derivative, \(f'(x)\), is given by applying the chain rule of trigonometric functions as follows:\n\n\(f'(x) = \cos(\beta x) * \beta\)\n\nThe second derivative, \(f''(x)\), is given by again applying the chain rule to the first derivative:\n\n\(f''(x) = -\sin(\beta x) * \beta^{2}\)\n\nFor the third derivative, \(f'''(x)\), apply the chain rule to the second derivative:\n\n\(f'''(x) = -\cos(\beta x) * \beta^{3}\)\n\nFinally, for the fourth derivative, \(f''''(x)\), apply the chain rule to the third derivative:\n\n\(f''''(x) = \sin(\beta x) * \beta^{4}\)

Verify the equation for the second derivative

We need to verify that the function \(f(x) = \sin(\beta x)\) and its second derivative satisfy the equation \(f''(x) + \beta^{2} * f(x) = 0\).\n\nSubstitute \(f(x) = \sin(\beta x)\) and the second derivative, \(f''(x)\), we found in Step 1 into the equation:\n\n\(-sin(\beta x) * \beta^{2} + \beta^{2} * sin(\beta x) = 0\)\n\nAfter simplifying, this verifies the provided equation, as the left side of the equation does result in 0.

Identify the general rules for the derivatives

Looking at the pattern of the derivatives game from Step 1, we can write general rules for the even and odd order derivatives.\n\nFor an even order derivative \(f^{(2k)}(x)\), the pattern repeats every four derivatives, so we alternate between \(\sin(\beta x)\) and \(-\sin(\beta x)\) due to the \((-1)\) factor. When k is even, \(f^{(2k)}(x) = \sin(\beta x) * \beta^{2k}\) and when k is odd, \(f^{(2k)}(x) = -\sin(\beta x) * \beta^{2k}\).\n\nFor an odd order derivative \(f^{(2k-1)}(x)\), the pattern also repeats every four derivatives, so we alternate between \(\cos(\beta x)\) and \(-\cos(\beta x)\). When k is even, \(f^{(2k-1)}(x) = \cos(\beta x) * \beta^{2k-1}\) and when k is odd, \(f^{(2k-1)}(x) = -\cos(\beta x) * \beta^{2k-1}\).

Key concepts

Chain Rule

The chain rule is a fundamental principle in calculus used to find the derivative of a composite function. It essentially states that if you have a function within a function, the derivative is found by multiplying the derivative of the outer function by the derivative of the inner function.

For example, for the sine function differentiation with respect to 'x', if we have a function like \(f(x) = \text{sin}(\beta x)\), where \(\beta\) is a constant and affects the input of the sine function, the chain rule would be applied as follows:

\[ f'(x) = \frac{d}{dx} \text{sin}(\beta x) = \cos(\beta x) \times \frac{d}{dx}(\beta x) = \cos(\beta x) \times \beta = \beta\cos(\beta x) \]
Understanding and applying the chain rule correctly is vital for accurately finding derivatives of trigonometric functions and for more complicated calculus problems.

Second Order Derivative

The second order derivative, represented as \( f''(x) \), provides information about the curvature or concavity of a function's graph. In other words, it tells us how the rate at which \( f'(x) \), the first derivative, changes.

For \( f(x) = \text{sin}(\beta x) \), the calculation of the second order derivative involves applying the differentiation process twice. Using the solution in our exercise:

\[ f''(x) = \frac{d}{dx}(\beta \cos(\beta x)) = -\beta^2 \sin(\beta x) \]
This second order derivative plays a critical role in verifying certain properties of the function, such as the equation \( f''(x) + \beta^2 f(x) = 0 \), which describes a harmonic oscillator in physics.

Higher Order Derivatives

Higher order derivatives refer to the derivatives beyond the first and second, involving the third derivative \(f'''(x)\), fourth derivative \(f''''(x)\), and so on. These derivatives can reveal patterns in a function's behavior and are essential in various applications, including Taylor series expansions and differential equations.

In determining higher order derivatives of \(f(x) = \sin(\beta x)\), one can see a pattern emerge: each derivative alternates between sine and cosine functions, multiplied by increasing powers of \(\beta\) and alternating signs determined by the order of the derivative:

\[ f^{(n)}(x) \] where 'n' is the order of the derivative.

Sine Function Differentiation

Differentiating the sine function involves understanding its basic derivative, which is the cosine function. When differentiating \(\sin(x)\), the result is \(\cos(x)\); when differentiating \(\sin(\beta x)\) with respect to 'x', such as in our exercise, you'll encounter the chain rule and multiply by \(\beta\) to obtain \(\beta\cos(\beta x)\).

The sine function differentiation is just the first step in a sequence that reveals a pattern as you calculate higher order derivatives. Notably, every fourth derivative lands back on the sine function, which plays into identifying the pattern in trigonometric derivatives.

Pattern in Derivatives

Identifying a pattern in the derivatives of trigonometric functions greatly simplifies the process of finding higher order derivatives. As seen in our example with \(f(x) = \sin(\beta x)\), a pattern repeats every four derivatives, switching between sine and cosine, and flipping signs due to the alternating powers of -1.

General rules can be formulated for even and odd derivatives: for even order \(f^{(2k)}(x)\) the pattern alternates based on the parity of 'k', while odd order \(f^{(2k-1)}(x)\) derivatives follow a similar pattern with cosine. These rules can be directly applied to simplify and expedite the calculation of derivatives in many trigonometric problems:

• Even order derivatives: \(f^{(2k)}(x) = (-1)^k \sin(\beta x) \beta^{2k}\)
• Odd order derivatives: \(f^{(2k-1)}(x) = (-1)^k \cos(\beta x) \beta^{2k-1}\)

By recognizing these patterns, students can efficiently tackle higher order differentiation without performing extensive calculations each time.