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Chapter 3: Problem 14

Exer. \(13-16:\) Find the first three derivatives. $$ f(x)=6 x^{4} $$

Short Answer

Expert verified
The first derivative is \(24x^3\), the second derivative is \(72x^2\), and the third derivative is \(144x\).

Step by step solution

01

Find the first derivative

To find the first derivative of the function, apply the power rule, which states that if \( f(x) = ax^n \), then the derivative \( f'(x) = nax^{n-1} \). Here, \( a = 6 \) and \( n = 4 \), so the first derivative is:\[ f'(x) = 4 \cdot 6 x^{4-1} = 24x^3. \]
02

Find the second derivative

Next, we find the second derivative by differentiating the first derivative. Again, use the power rule on \( f'(x) = 24x^3 \):\[ f''(x) = 3 \cdot 24 x^{3-1} = 72x^2. \]
03

Find the third derivative

Finally, differentiate the second derivative to find the third derivative. Apply the power rule to \( f''(x) = 72x^2 \):\[ f'''(x) = 2 \cdot 72 x^{2-1} = 144x. \]

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Key Concepts

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

Understanding the Power Rule
The power rule is a fundamental concept in calculus, particularly useful when dealing with derivatives. It provides a straightforward method for differentiating functions of the form \( f(x) = ax^n \). To use the power rule, follow these easy steps:
  • Identify the coefficient, \( a \), and the exponent, \( n \).
  • Multiply the coefficient by the exponent.
  • Reduce the exponent by one.

For example, if you're working with \( 6x^4 \), the power rule results in the derivative \( 4 \cdot 6x^{4-1} = 24x^3 \). This technique makes finding derivatives quick and efficient, especially for polynomials.
Finding the First Derivative
Derivatives represent the rate of change of a function. To find the first derivative, apply the power rule to the original function. Let's explore this with our sample function \( f(x) = 6x^4 \):
  • Apply the power rule: multiply the exponent \( 4 \) by the coefficient \( 6 \) to get \( 24 \).
  • Subtract one from the exponent \( 4 \) to result in \( 3 \).

The first derivative is \( f'(x) = 24x^3 \). This tells us how \( f(x) \) changes at any point \( x \), capturing its instantaneous rate of change.
Exploring the Second Derivative
The second derivative helps us understand the concavity and acceleration of a function. It is the derivative of the first derivative. Continuing from our example where the first derivative is \( f'(x) = 24x^3 \):
  • Again, apply the power rule: multiply the exponent \( 3 \) by \( 24 \) to yield \( 72 \).
  • Subtract one from the exponent \( 3 \) for a result of \( 2 \).

Thus, we find \( f''(x) = 72x^2 \). The second derivative provides insights into the function's curvature and whether it is curving upward or downward.
Understanding the Third Derivative
The third derivative is derived from the second derivative and is vital for understanding the rate at which concavity is changing. For our example, starting with the second derivative \( f''(x) = 72x^2 \):
  • Apply the power rule by multiplying the exponent \( 2 \) by \( 72 \) to get \( 144 \).
  • Reduce the exponent \( 2 \) by one, resulting in \( 1 \).

Therefore, the third derivative is \( f'''(x) = 144x \). This derivative can provide more detailed insights, such as the jerk or rate of change of acceleration, in practical applications.

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Most popular questions from this chapter

Two cars are approaching the same intersection along roads that run at right angles to each other. Car \(\mathrm{A}\) is traveling at \(20 \mathrm{mi} / \mathrm{hr}\), and car \(\mathrm{B}\) is traveling at \(40 \mathrm{mi} / \mathrm{hr}\). If, at a certain instant, \(\mathrm{A}\) is \(\frac{1}{4}\) mile from the intersection and \(\mathrm{B}\) is \(\frac{1}{2}\) mile from the intersection, find the rate at which they are approaching each other at that instant.

Find equations of the tangent line and the normal line to the graph of \(f\) at \(P\). $$ x^{2} y-y^{3}=8 ; \quad P(-3,1) $$

A point \(P\) moving on a coordinate line \(l\) has the given position function \(s\). When is its velocity 0 ? \(s(t)=t+2 \cos t\)

Shown is a graph of the function \(f\) with restricted domain. Find the points at which the tangent line is horizontal. \(f(x)=2 \sec x-\tan x ; \quad-\pi / 2

Graph \(f(x)=\frac{4}{16 \sin 2 x-x}\) on the interval [0,4] and estimate the \(x\) -coordinates of points at which the tangent line is horizontal.
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