Problem 40
Question
In Exercises, find the second derivative and solve the equation \(f^{\prime \prime}(x)=0\). $$ f(x)=x \sqrt{4-x^{2}} $$
Step-by-Step Solution
Verified Answer
The function's second derivative is \(f^{\prime\prime}(x)=-3x\). The value of \(x\) that makes the second derivative equal zero is \(x = 0\).
1Step 1: Finding the First Derivative
The derivative of the function \(f(x)=x \sqrt{4-x^{2}}\) is obtained by using the product and chain rule of differentiation. According to the product rule, the derivative of \(f(x)=u(x)\cdot v(x)\) is \(f'(x)=u'(x)v(x)+u(x)v'(x)\). Hence \(f'(x)=(1)\sqrt{4-x^{2}}+x(\frac{1}{2}(4-x^{2})^{-\frac{1}{2}} \cdot -2x)\) simplifying it results as \(f'(x)=\sqrt{4-x^{2}}-x^{2}\).
2Step 2: Finding the Second Derivative
The second derivative \(f^{\prime\prime}(x)\) is obtained by differentiating \(f'(x)\) using the chain rule again. The second derivative is: \(f^{\prime\prime}(x)=-(\frac{1}{2}(4-x^{2})^{-\frac{1}{2}} \cdot -2x)-2x-(2x)\), which simplifies to \(f^{\prime\prime}(x)=-3x\).
3Step 3: Solve for \(f^{\prime\prime}(x)=0\)
The equation \(f^{\prime\prime}(x)=0\) simplifies to -3x = 0. Solving for \(x\) gives \(x = 0\). This is the point where the function's second derivative equals zero.
Key Concepts
Product RuleChain RuleDifferentiation
Product Rule
In calculus, the Product Rule is a fundamental technique used for finding derivatives. It is essential when dealing with functions that are products of two differentiable functions. For a function defined as the product of two functions, say \( u(x) \) and \( v(x) \), the derivative is not simply the product of their individual derivatives. Instead, the Product Rule provides the formula:
In our exercise, we apply the Product Rule to the function \( f(x) = x \sqrt{4-x^2} \). Here, \( u(x) = x \) and \( v(x) = \sqrt{4-x^2} \). By differentiating \( u(x) \), we get \( u'(x) = 1 \). For \( v(x) \), using the Chain Rule, we find another component needed for differentiation, but more on that in the next section.
Understanding the Product Rule is crucial because it allows us to handle more complex differentiations, paving the way to solve problems involving combinations of simple functions. Through our original exercise, we identified and computed the first derivative using this powerful rule to reveal the changes in our function's behavior.
- \( f'(x) = u'(x)v(x) + u(x)v'(x) \)
In our exercise, we apply the Product Rule to the function \( f(x) = x \sqrt{4-x^2} \). Here, \( u(x) = x \) and \( v(x) = \sqrt{4-x^2} \). By differentiating \( u(x) \), we get \( u'(x) = 1 \). For \( v(x) \), using the Chain Rule, we find another component needed for differentiation, but more on that in the next section.
Understanding the Product Rule is crucial because it allows us to handle more complex differentiations, paving the way to solve problems involving combinations of simple functions. Through our original exercise, we identified and computed the first derivative using this powerful rule to reveal the changes in our function's behavior.
Chain Rule
The Chain Rule is another essential differentiation tool in calculus. It is used when differentiating composite functions — that is, functions applied one after the other. Essentially, if you have a function \( f(g(x)) \), the Chain Rule states:
In our given function \( f(x) = x \sqrt{4-x^2} \), the term \( \sqrt{4-x^2} \) demonstrates a classic example of where the Chain Rule is necessary. Here, the inner function \( g(x) = 4-x^2 \) is simplified under a square root, the outer function. We differentiate \( g(x) \) to get \( g'(x) = -2x \), which is then multiplied by the derivative of the square root function, according to the Chain Rule.
Utilizing the Chain Rule equips students with the ability to dissect and understand parts of more complicated functions. This rule, paired with the Product Rule, helped unfold our original equation, allowing us to solve for derivatives accurately and effectively.
- \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \)
In our given function \( f(x) = x \sqrt{4-x^2} \), the term \( \sqrt{4-x^2} \) demonstrates a classic example of where the Chain Rule is necessary. Here, the inner function \( g(x) = 4-x^2 \) is simplified under a square root, the outer function. We differentiate \( g(x) \) to get \( g'(x) = -2x \), which is then multiplied by the derivative of the square root function, according to the Chain Rule.
Utilizing the Chain Rule equips students with the ability to dissect and understand parts of more complicated functions. This rule, paired with the Product Rule, helped unfold our original equation, allowing us to solve for derivatives accurately and effectively.
Differentiation
Differentiation is the core process in calculus that involves finding the derivative of a function. The derivative represents the rate of change or the slope of the function at any given point. It allows us to analyze how a function behaves, which is crucial in various fields such as physics, economics, and engineering.
Derivatives are found through different rules, with our focus here being on the Product and Chain Rules. Both rules are tools of differentiation, extending basic principles to more complex, real-world applications. In our exercise, discovering the first and second derivatives of the function \( f(x) = x \sqrt{4-x^2} \) requires mastering these rules. The correct application reveals the behavior of the function, identifying critical points such as where the slope is zero or where changes in curvature occur.
Differentiation is powerful because it transforms abstract functions into tangible insights about their behavior. Using differentiation techniques, one can easily track and predict changes, offering solutions and interpretations for a wide array of practical problems.
Derivatives are found through different rules, with our focus here being on the Product and Chain Rules. Both rules are tools of differentiation, extending basic principles to more complex, real-world applications. In our exercise, discovering the first and second derivatives of the function \( f(x) = x \sqrt{4-x^2} \) requires mastering these rules. The correct application reveals the behavior of the function, identifying critical points such as where the slope is zero or where changes in curvature occur.
Differentiation is powerful because it transforms abstract functions into tangible insights about their behavior. Using differentiation techniques, one can easily track and predict changes, offering solutions and interpretations for a wide array of practical problems.
Other exercises in this chapter
Problem 40
In Exercises, use a graphing utility to graph the function and identify all relative extrema and points of inflection. $$ f(x)=x^{3}-3 x $$
View solution Problem 40
In Exercises, find the absolute extrema of the function on the interval \([0, \infty)\). $$ f(x)=8-\frac{4 x}{x^{2}+1} $$
View solution Problem 41
In Exercises, use a graphing utility to graph the function and identify all relative extrema and points of inflection. $$ f(x)=x^{3}-6 x^{2}+12 x $$
View solution Problem 41
The number \(y\) of medical degrees conferred in the United States from 1970 through 2004 can be modeled by \(y=0.813 t^{3}-55.70 t^{2}+1185.2 t+7752, \quad 0 \
View solution