Question

Prove that \(\frac{d^{2 n}}{d x^{2 n}}(\sin x)=(-1)^{n} \sin x \quad\) and \(\quad \frac{d^{2 n}}{d x^{2 n}}(\cos x)=(-1)^{n} \cos x\).

Short answer

Question: Prove the following relations for the nth derivative of sin(x) and cos(x): 1. \(\frac{d^{2n}}{dx^{2n}} (\sin x) = (-1)^n \sin x\) 2. \(\frac{d^{2n}}{dx^{2n}} (\cos x) = (-1)^n \cos x\) Answer: Following a step-by-step solution that analyzed the derivatives of sin(x) and cos(x), we identified a pattern and derived general formulas for the nth derivatives. By comparing these general formulas to the given relations, we have proved that the relations hold true for the nth derivative of sin(x) and cos(x).

Step-by-step solution

Find the first few derivatives of \(\sin x\) and \(\cos x\)

We'll start by finding the first four derivatives of \(\sin x\) and \(\cos x\): 1. \(\frac{d}{dx} (\sin x) = \cos x\) 2. \(\frac{d^2}{dx^2} (\sin x) = -\sin x\) 3. \(\frac{d^3}{dx^3} (\sin x) = -\cos x\) 4. \(\frac{d^4}{dx^4} (\sin x) = \sin x\) --- 1. \(\frac{d}{dx} (\cos x) = -\sin x\) 2. \(\frac{d^2}{dx^2} (\cos x) = -\cos x\) 3. \(\frac{d^3}{dx^3} (\cos x) = \sin x\) 4. \(\frac{d^4}{dx^4} (\cos x) = \cos x\)

Identify the pattern in the derivatives

From Step 1, we can observe the following pattern in the derivatives of \(\sin x\) and \(\cos x\): - Every second derivative of \(\sin x\) and \(\cos x\) is equal to their original functions (i.e., \(\sin x\) or \(\cos x\)), but with a potential change in sign. - The sign of the second derivatives alternate between positive and negative.

Find the general formula for the nth derivative

Now that we've identified the pattern, we can write the general formula for the nth derivative of \(\sin x\) and \(\cos x\): - \(\frac{d^{2n}}{dx^{2n}} (\sin x) = (-1)^n \sin x\) - \(\frac{d^{2n}}{dx^{2n}} (\cos x) = (-1)^n \cos x\)

Prove the given relations using the general formula

Using the general formulas we derived in Step 3, we can prove the given relations: 1. \(\frac{d^{2n}}{dx^{2n}} (\sin x) = (-1)^n \sin x\): This relation is already satisfied by the general formula for the nth derivative of \(\sin x\). 2. \(\frac{d^{2n}}{dx^{2n}} (\cos x) = (-1)^n \cos x\): This relation is also satisfied by the general formula for the nth derivative of \(\cos x\). Hence, we've proved that the given relations hold true for the nth derivative of \(\sin x\) and \(\cos x\).

Key concepts

Higher-Order Derivatives

Understanding higher-order derivatives is crucial for grasping the complexities of calculus. These derivatives are essentially the rates of change of rates of change. In simpler terms, if the first derivative represents the rate of change of a function, the second derivative represents the rate of change of the first derivative, and so on. This process can be repeated n times to obtain the nth derivative, often written as \( \frac{d^n}{dx^n} f(x) \).

For example, if the acceleration of an object is the second derivative of its position with respect to time, the third derivative might represent the jerk or the rate of change of acceleration. In trigonometric functions like \(\sin x\) and \(\cos x\), higher-order derivatives follow a cyclical pattern that simplifies their study and application in various fields like physics, engineering, and other sciences, where wave motion and oscillations are considered.

Sin and Cos Functions

The functions \(\sin x\) and \(\cos x\) are fundamental to trigonometry and calculus. These periodic functions have a range of -1 to 1 and are defined for all real numbers x. The function \(\sin x\) represents the y-coordinate of a point on the unit circle at an angle x from the positive x-axis, while \(\cos x\) represents the x-coordinate. Their derivatives are particularly noteworthy because of their repetitive nature. As we derive them, each derivative cycles through four possible outcomes: \(\sin x\), \(\cos x\), \(\-sin x\), and \(\-cos x\), in a continuous loop. This pattern is a key attribute that makes calculating higher-order derivatives of trigonometric functions more predictable than those of most other functions.

Calculus Proofs

Calculus proofs serve as a structured way to verify the truths within calculus, often using logical reasoning and mathematical principles. Proving the nth derivative formulas for \(\sin x\) and \(\cos x\) employs a methodical approach, as seen in the exercise. We discern patterns from the initial derivatives and then define a general formula that holds for all natural numbers n. This method not only allows us to understand the behavior of these functions but also to predict the outcome of their derivatives for any value of n.

In the context of the exercise, proving the nth derivative for these trigonometric functions helps students to solidify their comprehension of derivative operations, improving their ability to engage with more advanced calculus problems and concepts.

Derivative Patterns

Recognizing derivative patterns can greatly simplify the process of differentiation, especially when dealing with higher-order derivatives. In trigonometric functions, the derivatives of \(\sin x\) and \(\cos x\) exhibit a predictable cyclical pattern that repeats every four derivatives. As demonstrated in the problem-solving exercise, the pattern dictates that for every multiple of two derivatives of \(\sin x\) or \(\cos x\), we find ourselves back at the original function \(\sin x\) or \(\cos x\), with only the sign potentially changing. This alternating sign follows a sequence dependent on the exponent 2n, aligning with the power of (-1) raised to n. Such patterns are not just mathematical curiosities; they have practical implications in areas like signal processing, where knowing the nth derivative without extensive computation can be incredibly useful.