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Use the Chain Rule to find the derivative of the following functions. $$y=\left(1-e^{x}\right)^{4}$$

Short Answer

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Question: Find the derivative of the function $$y = (1-e^x)^4$$. Answer: The derivative of the given function is $$\frac{dy}{dx} = -4e^x(1-e^x)^3$$.

Step by step solution

01

Identify the outer and inner functions

Outer function: $$y(u)=u^4$$ Inner function: $$u(x)=1-e^x$$
02

Differentiate the outer function with respect to u

Differentiate the outer function with respect to u: $$\frac{dy}{du} = 4u^{3}$$
03

Differentiate the inner function with respect to x

Differentiate the inner function with respect to x: $$\frac{du}{dx} = -e^x$$
04

Apply the Chain Rule

Now, apply the Chain Rule by multiplying the derivative of the outer function by the derivative of the inner function: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 4u^3 \cdot (-e^x)$$
05

Substitute the inner function

Replace u with the inner function $$u(x) = 1 - e^x$$: $$\frac{dy}{dx} = 4(1-e^x)^3 \cdot (-e^x)$$ The derivative of the function is: $$\frac{dy}{dx} = -4e^x(1-e^x)^3$$

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Key Concepts

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

Derivative
A derivative is a fundamental concept in calculus that measures how a function changes as its input changes. Think of it as a rate of change or the slope of a function at a particular point. When you compute the derivative of a function, you are essentially finding how fast the function’s output is changing relative to changes in its input.

For example, in the given exercise, we are finding the derivative of the function \(y = (1-e^x)^4\). This means we want to know how \(y\) changes as \(x\) changes. By doing this, we gain insights into the behavior and shape of the function's graph, helping to identify points such as maxima, minima, and points of inflection.
Outer function
The outer function is the part of a composite function that forms the external operation. In the context of the Chain Rule, the outer function processes the intermediate result of the inner function. In our exercise, the outer function is \(y(u) = u^4\).

Whenever you're using the Chain Rule, it's essential to correctly identify this outer function because it plays a crucial role in differentiating composite functions:
  • It determines the overall structure of the operation you start with, before the inner function is plugged into it.
  • When differentiating, you initially take the derivative of the outer function with respect to its input (in this case, \(u\)).
  • This derivative helps simplify how the inner function will be further processed.
Understanding how to find and work with the outer function is pivotal to applying the Chain Rule successfully.
Inner function
The inner function is the component of a composite function that first acts on the variable. It's the first operation in the sequence before the outer function does its work. In the given exercise, the inner function is \(u(x) = 1-e^x\).

This piece is vital for several reasons:
  • It forms the bridge between the variable \(x\) and the outer operation, delivering the variable input that the outer function will process.
  • By differentiating the inner function, you determine how small changes in \(x\) affect its outcome, which is crucial for applying the Chain Rule.
  • In our context, understanding \(1-e^x\) allows us to see how it transforms \(x\) before the outer function works on it.
The role of the inner function is foundational in building up to the correct derivative outcome through the Chain Rule's formula.
Differentiation
Differentiation is the process of finding the derivative of a function. It involves calculating the rate at which a function's value changes with respect to changes in its input. This process is not only critical in finding the slope at specific points but also analyzing functions more broadly.

To differentiate composite functions, we use the Chain Rule, which allows us to take the derivative of one function nested inside another. Here’s a simplified approach:
  • First, identify both the outer and inner functions.
  • Differentiate the outer function with respect to the intermediate variable, then the inner function with respect to its variable.
  • Finally, combine these derivatives using multiplication to find the overall derivative.
In the exercise, we begin by differentiating \(y(u) = u^4\) to obtain \(4u^3\), and then \(u(x) = 1-e^x\) to get \(-e^x\).

Applying the Chain Rule gives us the final derivative \(\frac{dy}{dx} = -4e^x(1-e^x)^3\). Differentiation, particularly using the Chain Rule, unlocks understanding of how variables intricately affect each other in composite functions.

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Most popular questions from this chapter

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$y^{2}-3 x y=2$$

a. Use derivatives to show that \(\tan ^{-1} \frac{2}{n^{2}}\) and \(\tan ^{-1}(n+1)-\tan ^{-1}(n-1)\) differ by a constant. b. Prove that \(\tan ^{-1} \frac{2}{n^{2}}=\tan ^{-1}(n+1)-\tan ^{-1}(n-1)\)

The bottom of a large theater screen is \(3 \mathrm{ft}\) above your eye level and the top of the screen is \(10 \mathrm{ft}\) above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of \(3 \mathrm{ft} / \mathrm{s}\) while looking at the screen. What is the rate of change of the viewing angle \(\theta\) when you are \(30 \mathrm{ft}\) from the wall on which the screen hangs, assuming the floor is horizontal (see figure)?

Find the following higher-order derivatives. $$\frac{d^{n}}{d x^{n}}\left(2^{x}\right)$$

Economists use production functions to describe how the output of a system varies with respect to another variable such as labor or capital. For example, the production function \(P(L)=200 L+10 L^{2}-L^{3}\) gives the output of a system as a function of the number of laborers \(L\). The average product \(A(L)\) is the average output per laborer when \(L\) laborers are working; that is \(A(L)=P(L) / L\). The marginal product \(M(L)\) is the approximate change in output when one additional laborer is added to \(L\) laborers; that is, \(M(L)=\frac{d P}{d L}\). a. For the given production function, compute and graph \(P, A,\) and \(M\). b. Suppose the peak of the average product curve occurs at \(L=L_{0},\) so that \(A^{\prime}\left(L_{0}\right)=0 .\) Show that for a general production function, \(M\left(L_{0}\right)=A\left(L_{0}\right)\).

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