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Find the derivative of each function. \(f(x)=\left(x^{2}+2\right)^{5}\)

Short Answer

Expert verified
The derivative of the function \(f(x) = (x^2 + 2)^5\) is: \(f'(x) = 10x(x^2 + 2)^4\)

Step by step solution

01

Identify the Inner and Outer Functions

In the given function, \(f(x) = (x^2 + 2)^5\), we have an inner function and an outer function. The inner function is \(g(x) = x^2 + 2\), and the outer function is \(h(g(x)) = g(x)^5\).
02

Find the Derivative of the Inner Function

To find the derivative of the inner function, \(g(x) = x^2 + 2\), we use the power rule. The derivative of \(x^2 + 2\) with respect to \(x\) is: \(g'(x) = \frac{d(x^2 + 2)}{dx} = 2x\)
03

Find the Derivative of the Outer Function

Now, we need to find the derivative of the outer function, \(h(g(x)) = g(x)^5\). However, first, we'll find the derivative of \(h(u) = u^5\) with respect to \(u\), where \(u = g(x)\). Using the power rule, we get: \(h'(u) = \frac{d(u^5)}{du} = 5u^4\)
04

Use the Chain Rule to Find the Derivative of the Given Function

Now, we apply the chain rule: \(f'(x) = h'(g(x)) \cdot g'(x)\). Substitute the expressions we obtained in Steps 2 and 3: \(f'(x) = 5(g(x))^4 \cdot 2x = 5(x^2 + 2)^4 \cdot 2x\)
05

Final Answer

The derivative of the function \(f(x) = (x^2 + 2)^5\) is: \(f'(x) = 10x(x^2 + 2)^4\)

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Key Concepts

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

Chain Rule
The chain rule is a fundamental principle in calculus used for finding the derivative of composite functions. In essence, it allows you to differentiate a function that is made up of one function inside another. To apply the chain rule, you need to identify the 'inner' function and the 'outer' function.

For example, if you have a function like \( f(x) = (x^2 + 2)^5 \), the inner function is \( g(x) = x^2 + 2 \) and the outer function can be written as \( h(g(x)) = g(x)^5 \). The derivative of \( f(x) \) is then found by taking the derivative of the outer function with respect to the inner function, and multiplying it by the derivative of the inner function with respect to \( x \). This process can be expressed by the formula \( f'(x) = h'(g(x)) \cdot g'(x) \).
Power Rule
The power rule is an easy-to-use tool in derivative calculus that helps you effortlessly find the derivative of a function that is a power of \( x \). It states that if you have a function of the form \( f(x) = x^n \) where \( n \) is any real number, the derivative of \( f \) with respect to \( x \) is \( f'(x) = nx^{n-1} \).

For instance, consider the function \( g(x) = x^2 + 2 \). The power rule instructs you to bring down the exponent as a multiplier (2 in this case) and reduce the original exponent by one, giving us \( g'(x) = 2x \) as the derivative of the \( x^2 \) term. The constant term \( 2 \) disappears as the derivative of any constant is zero.
Function Derivative
The derivative of a function represents the rate at which the function value changes with respect to changes in its input value. It is a crucial concept in calculus, providing insights into the instantaneous rate of change, akin to the slope of the tangent to the function's graph at a given point.

Finding the derivative, or differentiation, involves applying derivative rules like the power rule, the product rule, the quotient rule, or the chain rule. Derivatives are widely used for various applications, from physics to finance, anywhere the concept of a changing rate needs to be quantified.
Calculus Problems
Calculus problems often involve finding derivatives, integrals, limits, and understanding the behavior of functions. These problems can range from simple polynomial differentiation to complex real-world applications involving rate of change and accumulation. To solve calculus problems, it is essential to have a clear understanding of different calculus rules and theorems and to practice applying these on various functions. Tackling these problems often requires a step-by-step approach, breaking down complicated functions into simpler parts, and carefully applying the right rules like the chain rule for composite functions or the power rule for polynomials.

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Most popular questions from this chapter

Extend the product rule for differentiation to the following case involving the product of three differentiable functions: Let \(h(x)=u(x) v(x) w(x)\) and show that \(h^{\prime}(x)=\) \(u(x) v(x) w^{\prime}(x)+u(x) v^{\prime}(x) w(x)+u^{\prime}(x) v(x) w(x)\) Hint: Let \(f(x)=u(x) v(x), g(x)=w(x)\), and \(h(x)=f(x) g(x)\) and apply the product rule to the function \(h\).

The distance \(s\) (in feet) covered by a car \(t\) sec after starting from rest is given by $$ s=-t^{3}+8 t^{2}+20 t \quad(0 \leq t \leq 6) $$ Find a general expression for the car's acceleration at any time \(t(0 \leq t \leq 6)\). Show that the car is decelerating \(2 \frac{2}{3}\) sec after starting from rest.

Find the derivative of each function. \(f(x)=\frac{1}{\sqrt{2 x^{2}-1}}\)

Find the derivative of each function. \(f(t)=\left(t^{-1}-t^{-2}\right)^{3}\)

HoTEL OccuPANCY RATES The occupancy rate of the allsuite Wonderland Hotel, located near an amusement park, is given by the function $$ r(t)=\frac{10}{81} t^{3}-\frac{10}{3} t^{2}+\frac{200}{9} t+60 \quad(0 \leq t \leq 12) $$ where \(t\) is measured in months, with \(t=0\) corresponding to the beginning of January. Management has estimated that the monthly revenue (in thousands of dollars/month) is approximated by the function $$ R(r)=-\frac{3}{5000} r^{3}+\frac{9}{50} r^{2} \quad(0 \leq r \leq 100) $$ where \(r\) is the occupancy rate. a. Find an expression that gives the rate of change of Wonderland's occupancy rate with respect to time. b. Find an expression that gives the rate of change of Wonderland's monthly revenue with respect to the occupancy rate. c. What is the rate of change of Wonderland's monthly revenue with respect to time at the beginning of January? At the beginning of July? Hint: Use the chain rule to find \(R^{\prime}(r(0)) r^{\prime}(0)\) and \(R^{\prime}(r(6)) r^{\prime}(6)\).

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