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Calculate the derivatives of all orders: \(f^{\prime}(x), f^{\prime \prime}(x), f^{\prime \prime \prime}(x), f^{(4)}(x), \ldots, f^{(n)}(x), \ldots\) \(f(x)=(-2 x+1)^{3}\)

Short Answer

Expert verified
In summary, the derivatives of all orders for the function \(f(x) = (-2x + 1)^3\) are: \(f'(x) = -6(-2x + 1)^2\) \(f''(x) = 24(-2x + 1)\) \(f'''(x) = -48\) \(f^{(n)}(x) = 0\) for \(n \geq 4\).

Step by step solution

01

1. Apply chain rule and find the first derivative f'(x)

We will use the chain rule to find the first derivative. The chain rule states that if y = u^n, where u is a function of x, then: \[ \frac{dy}{dx} = n * u^{(n-1)} * \frac{du}{dx}\] Applying the chain rule for f'(x): \[f'(x) = 3(-2x + 1)^{2} \cdot \frac{d}{dx}(-2x + 1)\] Now, take the derivative of the inner function (-2x + 1): \[\frac{d}{dx}(-2x + 1) = -2\] Substitute it into the expression for f'(x): \[f'(x) = 3(-2x + 1)^{2} \cdot (-2) = -6(-2x + 1)^{2}\]
02

2. Apply chain rule and find the second derivative f''(x)

Using the chain rule again for f''(x): \[f''(x) = \frac{d}{dx}\left[-6(-2x + 1)^{2}\right] = -12( -2x + 1) \cdot (-2)\] So, \[f''(x) = 24(-2x + 1)\]
03

3. Apply chain rule and find the third derivative f'''(x)

Once more, using the chain rule for f'''(x): \[f'''(x) = \frac{d}{dx}\left[24(-2x + 1)\right] = 24 \cdot (-2)\] Then, \[f'''(x) = -48\]
04

4. Apply chain rule and find the fourth derivative f^(4)(x)

Since f'''(x) is a constant value, the fourth and all higher order derivatives will be 0: \[f^{(4)}(x) = \frac{d}{dx}\left[-48\right] = 0\]
05

5. The pattern for the derivatives of all orders

From the steps above, we can see the pattern for the derivatives of all orders: f(x) = (-2x + 1)^3 f'(x) = -6(-2x + 1)^2 f''(x) = 24(-2x + 1) f'''(x) = -48 f^(4)(x) = 0 f^(n)(x) = 0 for n >= 4

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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 a fundamental tool in calculus used for finding the derivative of a composite function. When a function can be expressed as a function within another function, the chain rule allows us to differentiate it methodically. For instance, if you have a function like \(f(g(x))\), where both \(f\) and \(g\) are functions of \(x\), the derivative \(f'(x)\) is found by multiplying the derivative of the outer function evaluated at the inner function \(f'(g(x))\) by the derivative of the inner function, \(g'(x)\).To illustrate, consider \(f(x)=(-2x+1)^{3}\). We identify the inner function as \(g(x)=-2x+1\) and the outer function as \(f(u)=u^{3}\) where \(u=g(x)\). Following the chain rule:\[f'(x) = 3(-2x + 1)^{2} \times (-2)\]This approach simplifies complex differentiation by breaking it down into more manageable parts. Understanding and applying the chain rule is essential for higher mathematics and for finding derivatives of all orders as seen in subsequent steps of the exercise.
Higher-Order Derivatives
Higher-order derivatives are simply the derivatives of the derivative. Starting with the first derivative \(f'(x)\), the second derivative \(f''(x)\) is the derivative of \(f'(x)\), and this pattern continues for each successive derivative. In practice, higher-order derivatives measure the rates of change of lower-order derivatives and they can provide important physical insights, such as acceleration which is the second derivative of position with respect to time.In the given exercise, higher-order derivatives are computed using the chain rule recursively. The second derivative is:\[f''(x) = 24(-2x + 1)\]The third derivative is a constant:\[f'''(x) = -48\]After this point, any further derivatives will be zero because the derivative of a constant is zero. The concept of higher-order derivatives is often used in Taylor series expansions, optimization problems, and in the study of motion.
Derivative Patterns
In calculus, certain functions demonstrate patterns in their successive derivatives, which can simplify the process of finding derivatives of any order. This pattern becomes especially clear when working with polynomial and power functions. As we differentiate each time, the exponent decreases by one, until it reaches zero. At that point, any further derivatives will be zero, as the derivative of a constant is zero.The exercise shows a clear pattern in its derivatives:
  • The exponent of the inner function decreases by one with each derivative.
  • Each derivative introduces an additional factor of \( -2 \), which comes from the derivative of the inner function \( -2x + 1 \).
  • After the third derivative, the derivatives are all zero since the function becomes a constant.
Recognizing such patterns not only speeds up the differentiation process but also aids in understanding the behavior of the function and its graph. It's important to look for and utilize these patterns, as they can be a powerful tool in solving calculus problems.
Calculus
Calculus is a branch of mathematics that studies how things change. It provides a framework for modeling systems in which there is change, and a way to deduce the predictions of such models. It consists of two complementary branches: differential calculus, which concerns the rate of change of quantities; and integral calculus, which focuses on the aggregation of quantities over an interval.In the context of the given exercise, differential calculus is used to find the derivatives of a function at any order. This process involves finding the rate at which one quantity changes with respect to another. For example, understanding the rates of change within a motion can inform us about velocity and acceleration, two very practical applications of derivatives in real-world situations.In essence, calculus enables us to undertake complex dynamic models that are essential across various fields such as physics, engineering, economics, and beyond. Thus, it is not only a pillar of modern mathematics but also an indispensable tool in numerous scientific disciplines.

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Most popular questions from this chapter

You have been hired as a marketing consultant to Big Book Publishing, Inc., and you have been approached to determine the best selling price for the hit calculus text by Whiner and Istanbul entitled Fun with Derivatives. You decide to make life easy and assume that the demand equation for Fun with Derivatives has the linear form \(q=m p+b\), where \(p\) is the price per book, \(q\) is the demand in annual sales, and \(m\) and \(b\) are certain constants you'll have to figure out. a. Your market studies reveal the following sales figures: when the price is set at \(\$ 50.00\) per book, the sales amount to 10,000 per year; when the price is set at \(\$ 80.00\) per book, the sales drop to 1000 per year. Use these data to calculate the demand equation. b. Now estimate the unit price that maximizes annual revenue and predict what Big Book Publishing, Inc.'s annual revenue will be at that price.

The demand and unit price for your store's checkered T-shirts are changing with time. Show that the percentage rate of change of revenue equals the sum of the percentage rates of change of price and demand. (The percentage rate of change of a quantity \(Q\) is \(\left.Q^{\prime}(t) / Q(t) .\right)\)

In a related rates problem, there is no limit to the number of changing quantities we can consider. Illustrate this by creating a related rates problem with four changing quantities.

Demand for your tie-dyed T-shirts is given by the formula $$q=500-100 p^{0.5}$$ where \(q\) is the number of T-shirts you can sell each month at a price of \(p\) dollars. If you currently sell T-shirts for \(\$ 15\) each and you raise your price by \(\$ 2\) per month, how fast will the demand drop? (Round your answer to the nearest whole number.)

Daily oil production by Pemex, Mexico's national oil company, can be approximated by \(q(t)=-0.022 t^{2}+0.2 t+2.9\) million barrels \(\quad(1 \leq t \leq 9)\) where \(t\) is time in years since the start of \(2000 .^{54}\) At the start of 2008 the price of oil was \(\$ 90\) per barrel and increasing at a rate of \(\$ 80\) per year. \(^{55}\) How fast was Pemex's oil (daily) revenue changing at that time?

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