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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 in part (a) to write general rules for the even- and odd- order derivatives \(f^{(2 k)}(x)\) and \(f^{(2 k-1)}(x)\) [Hint: \((-1)^{k}\) is positive if \(k\) is even and negative if \(k\) is odd. \(]\)

Short Answer

Expert verified
The first four derivatives are: \(f'(x) = \beta \cos(\beta x)\), \(f''(x) = -\beta^2 \sin(\beta x)\), \(f'''(x) = -\beta^3 \cos(\beta x)\), and \(f''''(x) = \beta^4 \sin(\beta x)\). The equation \(f''(x) + \beta^2 f(x) = 0\) has been verified. The general rules for even and odd parities are \(f^{(2k)}(x) = (\beta^{2k})^{(-1)^k} \sin(\beta x)\) and \(f^{(2k-1)}(x) = (\beta^{2k-1})^{(-1)^{k-1}} \cos(\beta x)\) respectively.

Step by step solution

01

Differentiation of the function

The first step is to find the first four derivatives of the function. The first derivative is obtained by applying the chain rule. \For \(f'(x)\), we use the chain rule to differentiate \( \sin(\beta x)\) which gives us \( \beta \cos(\beta x)\). \The second derivative, \(f''(x)\), is \(-\beta^2 \sin(\beta x)\), the derivative of the first derived function. \Applying the chain rule again, the third derivative, \(f'''(x)\), is \(-\beta^3 \cos(\beta x)\). \Finally, the fourth derivative, \(f''''(x)\), is \(\beta^4 \sin(\beta x)\), the derivative of the third derived function.
02

Verification of equation

Take the second derivative and add the result with the original function multiplied by \(\beta^2\), we get: \(-\beta^2 \sin(\beta x) + \beta^2 \sin(\beta x) = 0\). Thus verifying the equation \(f''(x) + \beta^2 f(x) = 0\).
03

General rules formulation

Looking at the results from the differentiation step, we notice a pattern forming with the derivatives of the function. For the even-order derivatives, the pattern is \(f^{(2k)}(x) = (\beta^{2k})^{(-1)^k} \sin(\beta x)\), while the pattern for the odd-order derivatives is \(f^{(2k-1)}(x) = (\beta^{2k-1})^{(-1)^{k-1}} \cos(\beta x)\). This is due to the sine function alternating between positive and negative after each differentiation, and after two differentiations, \(f(x)\) returns back to original function, just with a multiplier of \(-1\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is an essential technique in calculus differentiation used to differentiate composite functions. When you have a function nested within another, you can think of the chain rule as peeling back layers, like an onion. For example, in the function \(f(x) = \sin(\beta x)\), we have an outer function, \(\sin(u)\), and an inner function, \(u = \beta x\).
To differentiate this using the chain rule:
  • First, differentiate the outer function with respect to the inner function: \(\cos(u)\).
  • Then, differentiate the inner function with respect to \(x\): \(\beta\).
So applying this to our example, the first derivative \(f'(x)\) becomes \(\beta \cos(\beta x)\).
Notice how we're packaging two elements: the derivative of the outer and inner functions to find the derivative of the composite function. The chain rule shines in problems where we see compositions like trigonometric, exponential, or logarithmic functions with other expressions.
Higher-Order Derivatives
Higher-order derivatives refer to derivatives taken beyond the first. For the function \(f(x) = \sin(\beta x)\), by repeatedly applying calculus rules, we discover its behavior through the sequence of derivatives:
  • The first derivative is \(f'(x) = \beta \cos(\beta x)\).
  • The second derivative is \(f''(x) = -\beta^2 \sin(\beta x)\).
  • The third derivative is \(f'''(x) = -\beta^3 \cos(\beta x)\).
  • The fourth derivative is \(f''''(x) = \beta^4 \sin(\beta x)\).
This pattern of signs and trigonometric functions repetition is common.
Through observation, one can identify that every two differentiations return the function to sine and cosine with changed signs due to the periodic nature of trigonometric functions. Higher-order derivatives are crucial as they offer insights into the movement and curvature of a function, useful in physics and engineering.
Trigonometric Functions
Trigonometric functions like \(\sin(x)\) and \(\cos(x)\) are fundamental in calculus due to their cyclical nature. They repeatedly oscillate between 1 and -1, which influences their differentiability.
When differentiating trigonometric functions like \(\sin(\beta x)\), their periodic nature results in a recurring pattern:
  • The derivative of \(\sin(x)\) is \(\cos(x)\), and the derivative of \(\cos(x)\) is \(-\sin(x)\).
  • This results in an alternation of sine and cosine, with accompanying changes in sign.
The exercise above demonstrates that differentiating \(\sin(\beta x)\), employs these features, generating a predictable series.
Furthermore, the exercise confirms the specific relationship \(f''(x) + \beta^2 f(x) = 0\), an equation showcasing harmonic behavior. Recognizing such properties in trigonometric differentiation is vital since it simplifies the process and frequently appears in oscillatory motion and wave theory.

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