Problem 60
Question
Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ f(x)=x^{3}(x-4)^{2} $$
Step-by-Step Solution
Verified Answer
The derivative of the function \(f(x) = x^{3}(x-4)^{2}\) is \(f'(x) = 2x^{3}(x-4) + 3x^{2}(x-4)^{2}\). The differentiation rules used here are the product rule and the chain rule.
1Step 1: Identify the Functions to Apply the Product Rule
Here \(f(x)\) is a product of two functions, namely \(u = x^{3}\) and \(v = (x-4)^{2}\). Therefore, the product rule is applied first, that states the derivative of a product of two functions is the first function times the derivative of the second plus the second function times the derivative of the first.
2Step 2: Apply the Product Rule and Initiate Chain Rule
Start by differentiating \(u\) and \(v\) separately. The derivative of \(u\), \(u'= 3x^{2}\). For \(v\) it's a little bit complex because it's a function of a function, therefore the chain rule of differentiation is used here, both for differentiating and simplifying. \(v'\) is then calculated as \(2(x-4)*1 = 2(x-4)\).
3Step 3: Final Derivative Expression
Now using the product rule with \(u\), \(u'\), \(v\) and \(v'\), \(f'(x) = u*v' + v*u' = x^{3}*2(x-4) + (x-4)^{2}*3x^{2}=2x^{3}(x-4) + 3x^{2}(x-4)^{2}\). This is the derivative of the function \(f(x)\) which was sought.
Key Concepts
Derivative of a FunctionProduct RuleChain Rule
Derivative of a Function
Understanding the derivative of a function is foundational to calculus. In essence, it represents how a function changes as its input changes. Intuitively, it's like observing the speedometer of a car; it shows how fast the car's position is changing over time.
For any function, say, \(f(x)\), the derivative is typically denoted as \(f'(x)\) and is calculated using various rules of differentiation. These rules are shortcuts that make it easier to find the rate at which a function's output is changing at any given point without having to use the limit definition of the derivative every single time.
For example, the power rule is a simple rule used to differentiate functions in the form of \(x^n\), where \(n\) is any real number. According to the power rule, the derivative of such a function is \(nx^{n-1}\). So for \(f(x) = x^3\), applying the power rule gives us \(f'(x) = 3x^2\), which represents the instantaneous rate of change of the cubic function at any point \(x\).
For any function, say, \(f(x)\), the derivative is typically denoted as \(f'(x)\) and is calculated using various rules of differentiation. These rules are shortcuts that make it easier to find the rate at which a function's output is changing at any given point without having to use the limit definition of the derivative every single time.
For example, the power rule is a simple rule used to differentiate functions in the form of \(x^n\), where \(n\) is any real number. According to the power rule, the derivative of such a function is \(nx^{n-1}\). So for \(f(x) = x^3\), applying the power rule gives us \(f'(x) = 3x^2\), which represents the instantaneous rate of change of the cubic function at any point \(x\).
Product Rule
When a function is composed of two multiplied sub-functions, finding its derivative requires the product rule. It's like observing two gears working together in a machine; the product rule helps us understand how the overall system is functioning when both parts contribute to the motion.
The product rule states that the derivative of a product of two functions \(u(x)\) and \(v(x)\) is \(u'(x)v(x) + u(x)v'(x)\). You keep one function the same, differentiate the other, and then switch roles, finally adding both results together.
Using our exercise as an illustration, with \(u = x^3\) and \(v = (x-4)^2\), we apply the product rule to get the derivative of \(f(x)\). The result is the sum of the first function \(u\) times the derivative of the second \(v'\), plus the second function \(v\) times the derivative of the first \(u'\), which gives us \(f'(x) = u*v' + v*u'\).
The product rule states that the derivative of a product of two functions \(u(x)\) and \(v(x)\) is \(u'(x)v(x) + u(x)v'(x)\). You keep one function the same, differentiate the other, and then switch roles, finally adding both results together.
Using our exercise as an illustration, with \(u = x^3\) and \(v = (x-4)^2\), we apply the product rule to get the derivative of \(f(x)\). The result is the sum of the first function \(u\) times the derivative of the second \(v'\), plus the second function \(v\) times the derivative of the first \(u'\), which gives us \(f'(x) = u*v' + v*u'\).
Chain Rule
The chain rule assists us when dealing with composite functions, which are functions within functions, similar to stacking Russian dolls. It allows us to unwrap the layers and find the rate at which the entire set changes.
Explicitly, the chain rule states that to find the derivative of a composite function \(v(u(x))\), you multiply the derivative of the outer function \(v'(u)\) by the derivative of the inner function \(u'(x)\), resulting in \(v'(u)*u'(x)\).
In our exercise, \(v = (x-4)^2\) is a composite function (\(u\) being \(x-4\) and the outer function being \(u^2\)). To differentiate \(v\), we find the derivatives of the inner and outer layers separately. The derivative of the inner function \(u\) is \(u' = 1\) (since \(x-4\) is linear), and the derivative of the outer function with respect to \(u\) is \(v'(u) = 2u\), using the power rule. The chain rule then gives us \(v' = 2(u)*u' = 2(x-4)*1\), which is used in conjunction with the product rule to find \(f'(x)\).
By understanding how to apply these differentiation rules, students can tackle complex derivatives with confidence and precision.
Explicitly, the chain rule states that to find the derivative of a composite function \(v(u(x))\), you multiply the derivative of the outer function \(v'(u)\) by the derivative of the inner function \(u'(x)\), resulting in \(v'(u)*u'(x)\).
In our exercise, \(v = (x-4)^2\) is a composite function (\(u\) being \(x-4\) and the outer function being \(u^2\)). To differentiate \(v\), we find the derivatives of the inner and outer layers separately. The derivative of the inner function \(u\) is \(u' = 1\) (since \(x-4\) is linear), and the derivative of the outer function with respect to \(u\) is \(v'(u) = 2u\), using the power rule. The chain rule then gives us \(v' = 2(u)*u' = 2(x-4)*1\), which is used in conjunction with the product rule to find \(f'(x)\).
By understanding how to apply these differentiation rules, students can tackle complex derivatives with confidence and precision.
Other exercises in this chapter
Problem 59
Population Growth A population of bacteria is introduced into a culture. The number of bacteria \(P\) can be modeled by $$P=500\left(1+\frac{4 t}{50+t^{2}}\righ
View solution Problem 60
Revenue The revenue \(R\) (in millions of dollars per year) for Polo Ralph Lauren from 1999 through 2005 can be modeled by $$ \begin{aligned} R=0.59221 t^{4} &-
View solution Problem 60
Describe the \(x\) -values at which \(f\) is differentiable. $$ f(x)=\left\\{\begin{array}{ll}{x^{2}-3,} & {x \leq 0} \\ {3-x^{2},} & {x>0}\end{array}\right. $$
View solution Problem 60
Quality Control The percent \(P\) of defective parts produced by a new employee \(t\) days after the employee starts work can be modeled by $$P=\frac{t+1750}{50
View solution