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Differentiate the function. \(F(x)=\frac{3}{4} x^{8}\)

Short Answer

Expert verified
The derivative is \(F'(x) = 6x^7\).

Step by step solution

01

Understand the Power Rule

To differentiate the function, apply the power rule of differentiation, which states that if you have a function of the form \(f(x) = ax^n\), the derivative \(f'(x)\) is given by multiplying the exponent \(n\) by the coefficient \(a\) and then reducing the exponent by 1. Symbolically, this can be written as \(f'(x) = anx^{(n-1)}\).
02

Apply the Power Rule to the Function

Given the function \(F(x) = \frac{3}{4}x^8\), identify \(a = \frac{3}{4}\) and \(n = 8\). Apply the power rule to find the derivative: \[F'(x) = \frac{3}{4} \times 8 \times x^{8-1}\].
03

Simplify the Expression

Calculate \(\frac{3}{4} \times 8\), which equals 6. Then, write the derivative as: \[F'(x) = 6x^7\].

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Key Concepts

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

Power Rule
The power rule is a fundamental concept in calculus used to find derivatives of polynomial functions. It simplifies the process of differentiation by providing a straightforward method. The rule applies to any function of the form \(f(x) = ax^n\), where \(a\) is a constant and \(n\) is a positive integer. To differentiate such a function, you multiply the exponent \(n\) by the constant \(a\), and then subtract 1 from the exponent. This gives a new function, known as the derivative, expressed as \(f'(x) = anx^{n-1}\).

When applying the power rule, it’s essential to recognize the terms in your function: the coefficient \(a\) and the exponent \(n\). Once identified, plug these values into the power rule formula. This method is quick and practical, allowing you to differentiate expressions efficiently.
  • Identify the coefficient \(a\) and the exponent \(n\).
  • Multiply \(a\) by \(n\).
  • Reduce the exponent by one.

These steps are invaluable in terms of understanding and applying calculus to real-world problems.
Derivative of Polynomial Functions
Polynomial functions, which consist of terms of the form \(ax^n\), are common in calculus. Differentiating these functions allows us to discover the rate at which they change. This is crucial in various fields such as physics, engineering, and economics.

When dealing with polynomial functions, each term can be differentiated independently using the power rule. For example, if we consider a polynomial like \(F(x) = \, \frac{3}{4} x^8\), we only have one term, and thus apply the power rule directly to it. Changing the function using differentiation reveals the function's behavior, which is critical in understanding its application in various scenarios.
If your function had more terms, you would simply repeat the process for each term individually.
  • Differentiate each term separately using the power rule.
  • Combine the results to get the overall derivative.

This process highlights how powerful calculus can be in analyzing mathematical models.
Calculus
Calculus is a branch of mathematics that deals with rates of change and the accumulation of quantities. It is divided primarily into two areas: differentiation and integration. Differentiation, as seen with the power rule and polynomial functions, focuses on how functions change. This is done by finding their derivatives, providing insights into the function's slope and behavior at various points.

Differentiation helps understand how varying one quantity affects another, a principle widely used in scientific and engineering computations, such as predicting speeds or determining the behavior of dynamic systems.
  • Key areas of calculus include differentiation and integration.
  • It allows for the analysis of changing systems.
  • Used in various disciplines for modeling and interpretation.

Grasping these concepts in calculus is vital as they form the foundation for more advanced studies and practical applications in fields like biology, computer science, and economics.

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Most popular questions from this chapter

On page 431 of Physics: Calculus, 2 \(\mathrm{d}\) ed. by Eugene Hecht (Pacific Grove, \(\mathrm{CA}, 2000 ),\) in the course of deriving the formula \(T=2 \pi \sqrt{L / g}\) for the period of a pendulum of length \(L,\) the author obtains the equation \(a_{T}=-g \sin \theta\) for the tangential acceleration of the bob of the pendulum. He then says, "for small angles, the value of \(\theta\) in radians is very nearly the value of \(\sin \theta ;\) they differ by less than 2\(\%\) out to about \(20^{\circ} . "\) (a) Verify the linear approximation at 0 for the sine function: \(\sin x \approx x\) (b) Use a graphing device to determine the values of \(x\) for which sin \(x\) and \(x\) differ by less than 2\(\%\) . Then verify Hecht's statement by converting from radians to degrees.

Two sides of a triangle have lengths 12 \(\mathrm{m}\) and 15 \(\mathrm{m}\) . The angle between them is increasing at a rate of 2\(\% / \mathrm{min}\) . How fast is the length of the third side increasing when the angle between the sides of fixed length is \(60^{\circ} ?\)

Find the value of the number \(a\) such that the families of curves \(y=(x+c)^{-1}\) and \(y=a(x+k)^{1 / 3}\) are orthogonal trajectories.

A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 600 \(\mathrm{ft} / \mathrm{s}\) when it has risen 3000 \(\mathrm{ft.}\) (a) How fast is the distance from the television camera to the rocket changing at that moment? (b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

Explain, in terms of linear approximations or differentials, why the approximation is reasonable. $$\sec 0.08 \approx 1$$

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