Problem 34
Question
Find \(y^{(4)}=d^{4} y / d x^{4}\) if a. \(y=-2 \sin x\) b. \(y=9 \cos x\)
Step-by-Step Solution
Verified Answer
(a) \(y^{(4)} = -2 \sin x\); (b) \(y^{(4)} = 9 \cos x\)
1Step 1: Differentiate the function for the first time (a)
For the function \(y = -2 \sin x\), the first derivative is \(y' = \frac{d(-2 \sin x)}{dx} = -2 \cos x\).
2Step 2: Differentiate the function a second time (a)
The second derivative is \(y'' = \frac{d(-2 \cos x)}{dx} = 2 \sin x\).
3Step 3: Differentiate the function a third time (a)
The third derivative is \(y''' = \frac{d(2 \sin x)}{dx} = 2 \cos x\).
4Step 4: Differentiate the function a fourth time (a)
For the fourth derivative, we find \(y^{(4)} = \frac{d(2 \cos x)}{dx} = -2 \sin x\).
5Step 1: Differentiate the function for the first time (b)
For the function \(y = 9 \cos x\), the first derivative is \(y' = \frac{d(9 \cos x)}{dx} = -9 \sin x\).
6Step 2: Differentiate the function a second time (b)
The second derivative is \(y'' = \frac{d(-9 \sin x)}{dx} = -9 \cos x\).
7Step 3: Differentiate the function a third time (b)
The third derivative is \(y''' = \frac{d(-9 \cos x)}{dx} = 9 \sin x\).
8Step 4: Differentiate the function a fourth time (b)
The fourth derivative is \(y^{(4)} = \frac{d(9 \sin x)}{dx} = 9 \cos x\).
Key Concepts
Fourth DerivativeTrigonometric FunctionsDifferentiation Steps
Fourth Derivative
In calculus, finding the fourth derivative of a function involves differentiating the function four times. The process of differentiation is about finding the rate at which a function changes. In this particular exercise, we worked on two trigonometric functions:
This recurrence happens because the derivatives of sine and cosine functions alternate between each other with a sign change, leading to a predictable pattern.
- For the function \(y = -2 \sin x\), the journey of differentiation takes us back to the original function after four steps. We started with \(-2 \cos x\), then \(2 \sin x\), followed by \(2 \cos x\), and finally returned to \(-2 \sin x\).
- Similarly, for \(y = 9 \cos x\), we moved through \(-9 \sin x\), \(-9 \cos x\), \(9 \sin x\), and ended up back to \(9 \cos x\).
This recurrence happens because the derivatives of sine and cosine functions alternate between each other with a sign change, leading to a predictable pattern.
Trigonometric Functions
Trigonometric functions such as sine and cosine are fundamental in calculus due to their repetitive oscillating nature. These functions have several distinct properties:
- **Sine Function (\(\sin x\))**: When differentiated for the first time, the result is the cosine function. Each subsequent derivative continues a predictable pattern: sine, cosine, negative sine, negative cosine, and so on.
- **Cosine Function (\(\cos x\))**: The cosine function follows a similar pattern, with derivatives flipping between negative and positive while alternating between cosine and sine functions.
Differentiation Steps
The differentiation process involves sequentially finding the derivatives of a function. Let's break down these steps:
- **First Differentiation**: When differentiating \(y = -2 \sin x\), the first derivative is \(-2 \cos x\). For \(y = 9 \cos x\), the first derivative is \(-9 \sin x\). This involves applying basic differentiation rules known as chain rules and derivatives of trigonometric functions.
- **Second Differentiation**: We continue by taking the derivative of the result from the first step. This results in switching the function from cosine to sine, or vice versa, and sometimes introduces a negative sign.
- **Third and Fourth Differentiation**: Continuing this pattern leads us to a third and then finally back to our original expression at the fourth integration.
Other exercises in this chapter
Problem 34
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Find the derivatives of the function. $$y=x^{-3 / 3}+\pi^{3 / 2}$$
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