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Chapter 4: Problem 73

Find the first and the second derivatives of each function. \(f(x)=x^{3}-3 x^{2}+1\)

Short Answer

Expert verified
The first derivative is \(f'(x) = 3x^2 - 6x\), and the second derivative is \(f''(x) = 6x - 6\).

Step by step solution

01

Find the first derivative

To find the first derivative of the function, we'll differentiate each term separately. The function is given by:\[f(x) = x^3 - 3x^2 + 1\]The first derivative \(f'(x)\) is obtained by applying the power rule (\(\frac{d}{dx}[x^n] = nx^{n-1}\)) to each term:- For \(x^3\), the derivative is \(3x^2\).- For \(-3x^2\), the derivative is \(-6x\).- The derivative of a constant, \(1\), is \(0\).So, the first derivative \(f'(x)\) of the function is:\[f'(x) = 3x^2 - 6x\]
02

Find the second derivative

To find the second derivative, we differentiate the first derivative \(f'(x)\). Given:\[f'(x) = 3x^2 - 6x\]We'll apply the power rule again:- For \(3x^2\), the derivative is \(6x\).- For \(-6x\), the derivative is \(-6\).Therefore, the second derivative \(f''(x)\) is:\[f''(x) = 6x - 6\]

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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 the derivative of functions that are expressed as power functions. The rule is stated as: if you have a function in the form of \(x^n\), then its derivative is \(nx^{n-1}\). This simple yet powerful rule allows you to quickly calculate derivatives without complicated calculations.
For instance, when you have a term like \(x^3\):
  • The exponent \(n\) is 3.
  • According to the power rule, take the exponent, multiply it by the coefficient (which is 1 for \(x^3\) since no other number is multiplying \(x^3\)), and then reduce the exponent by 1.
  • This results in a derivative of \(3x^2\).
The power rule simplifies the process of differentiation, especially for polynomial functions where each term can be treated separately. Memorizing this rule makes it easier to tackle more complex calculus problems.
First Derivative
The first derivative of a function helps to determine the rate at which the function's value is changing. It can be interpreted as the slope of the tangent line to the function at any given point.
By differentiating the function \(f(x)=x^3-3x^2+1\) using the power rule, you can find the first derivative \(f'(x)\):
  • For \(x^3\), applying the power rule gives you \(3x^2\).
  • For \(-3x^2\), the first derivative is \(-6x\).
  • The constant \(1\) becomes \(0\) when differentiated, as constants do not change.
These steps lead to the first derivative: \(f'(x) = 3x^2 - 6x\). Understanding the first derivative is crucial as it provides insights into increasing or decreasing behavior of the function across its domain.
Second Derivative
The second derivative provides information on the concavity of a function and can indicate points of inflection. It is the derivative of the first derivative and gives a deeper insight into the function's behavior.
You find the second derivative by differentiating the first derivative \(f'(x) = 3x^2 - 6x\):
  • For \(3x^2\), the derivative is \(6x\) using the power rule.
  • For \(-6x\), the derivative is simply \(-6\), as the derivative of \(x\) is 1.
Therefore, the second derivative is \(f''(x) = 6x - 6\). This second derivative can tell you where the graph of the function is concave up or concave down, influencing how steeply the graph increases or decreases.

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Most popular questions from this chapter

Compute \(f(c+h)-f(c)\) at the indicated point. $$ f(x)=\frac{1}{x} ; c=-2 $$

A measurement error in \(x\) affects the accuracy of the value \(f(x) .\) In each case, determine an interval of the form $$[f(x)-\Delta f, f(x)+\Delta f]$$ that reflects the measurement error \(\Delta x .\) In each problem, the quantities given are \(f(x)\) and \(x=\) true value of \(x \pm|\Delta x|\). $$ f(x)=1-3 x, x=-2 \pm 0.3 $$

Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ $$ f(x)=\left(1+\frac{1}{x}\right)^{1 / 4} \text { at } a=1 $$

Assume that \(N(t)\) denotes the size of a population at time \(t\) and that \(N(t)\) satisfies the differential equation $$ \frac{d N}{d t}=3 N\left(1-\frac{N}{20}\right) $$ Let \(f(N)=3 N\left(1-\frac{N}{20}\right)\) for \(N \geq 0\). Graph \(f(N)\) as a function of \(N\) and identify all equilibria (i.e., all points where \(\frac{d N}{d t}=0\) ).

Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ $$ f(x)=\frac{2}{1+x} \text { at } a=1 $$
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