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Calculate the derivatives of all orders: \(f^{\prime}(x), f^{\prime \prime}(x), f^{\prime \prime \prime}(x), f^{(4)}(x), \ldots, f^{(n)}(x), \ldots\) . \(f(x)=4 x^{2}-x+1\)

Short Answer

Expert verified
The derivatives of all orders for the function \(f(x) = 4x^2 - x + 1\) are: \[f^{(n)}(x) = \begin{cases} 8x - 1 & \text{for } n = 1 \newline 8 & \text{for } n = 2 \newline 0 & \text{for } n \geq 3 \end{cases}\]

Step by step solution

01

Calculate the first derivative (f'(x))

To calculate the first derivative, we'll apply the power rule: \(\frac{d}{dx}(x^n) = nx^{(n-1)}\). So, differentiating the given function \(f(x) = 4x^2 - x + 1\) with respect to \(x\): \[f'(x) = \frac{d}{dx}(4x^2) - \frac{d}{dx}(x) + \frac{d}{dx}(1)\] \[f'(x) = 8x - 1\]
02

Calculate the second derivative (f''(x))

Now, we will calculate the second derivative of the function by differentiating \(f'(x)\) with respect to \(x\): \[f''(x) = \frac{d}{dx}(8x - 1)\] \[f''(x) = 8\]
03

Calculate the third derivative (f'''(x))

Next, we will calculate the third derivative of the function by differentiating \(f''(x)\) with respect to \(x\): \[f'''(x) = \frac{d}{dx}(8)\] \[f'''(x) = 0\]
04

Identify the pattern and the general formula for the nth derivative

We see that after the third derivative, we are differentiating a constant. The derivative of a constant is always 0. This means that all higher-order derivatives of \(f(x)\) will also be 0. Thus, we can conclude: \[f^{(n)}(x) = \begin{cases} 8x - 1 & \text{for } n = 1 \newline 8 & \text{for } n = 2 \newline 0 & \text{for } n \geq 3 \end{cases}\] So, we have found the derivatives of all orders for the given function \(f(x) = 4x^2 - x + 1\).

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Key Concepts

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

Higher-Order Derivatives
When we talk about derivatives, we usually refer to the slope of a curve at a particular point. But what if we want to go beyond just the first derivative?Higher-order derivatives involve taking the derivative of a derivative.This means you continue differentiating the expression until you achieve the order you need.Let's take the example of the function given: \[ f(x) = 4x^2 - x + 1 \]- **First Derivative (\(f'(x)\)):** This involves applying the basic derivative rules to get \[ f'(x) = 8x - 1 \] This represents the rate of change of the original function. - **Second Derivative (\(f''(x)\)):** By differentiating \(f'(x)\), we obtain: \[ f''(x) = 8 \] It represents the rate of change of the rate of change. - **Third Derivative (\(f'''(x)\)) and Beyond:** Continuing, we differentiate \(f''(x)\) to get: \[ f'''(x) = 0 \] Any subsequent derivatives would also be 0 since we are now differentiating a constant.
Power Rule
The power rule is one of the most important techniques to know when it comes to differentiation.It tells us how to differentiate expressions of the form \(x^n\).The power rule can be stated as:\[ \frac{d}{dx}(x^n) = nx^{n-1} \]Let's see how the power rule applies in the function we have:- For the term \(4x^2\): - According to the power rule, differentiating \(4x^2\) gives: \[ 4 imes 2x^{2-1} = 8x \] - For the term \(-x\): - Since \(x\) is equivalent to \(x^1\), the power rule gives us: \[ 1 imes x^{1-1} = 1 \] Combining these, you obtain the first derivative:\[ f'(x) = 8x - 1 \]Understanding the power rule makes finding derivatives simpler and more efficient.
Mathematical Patterns
Identifying patterns in mathematical operations can significantly ease the process of solving problems.When dealing with derivatives, establishing a pattern allows for predicting future calculations.In the example:- We noticed that after calculating the third derivative of \(f(x)\), we obtain a constant. - At this point, any further derivatives would simply result in 0. - This is because the derivative of a constant is always 0.The pattern helps in finding derivatives of all orders without recalculating each one individually:\[ f^{(n)}(x) =\begin{cases} 8x - 1 & \text{for } n = 1 \ 8 & \text{for } n = 2 \ 0 & \text{for } n \geq 3\end{cases}\]Recognizing these patterns not only saves time but also builds a deeper understanding of the differentiation process.

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Most popular questions from this chapter

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