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Chapter 3: Problem 48

Compute the derivative of the following functions. $$f(x)=(1-2 x) e^{-x}$$

Short Answer

Expert verified
Answer: The derivative of the function is \(f'(x) = (1-2x)e^{-x} - 2e^{-x}\).

Step by step solution

01

Find the derivative of u(x)

First, let's find the derivative of u(x): $$u'(x) = \frac{d}{dx}(1-2x)$$ Using the basic differentiation rules, we get: $$u'(x) = -2$$
02

Find the derivative of v(x)

Next, let's find the derivative of v(x): $$v'(x) = \frac{d}{dx}(e^{-x})$$ Using the chain rule, we get: $$v'(x) = -e^{-x}$$
03

Apply the product rule

Now that we have u'(x) and v'(x), let's apply the product rule to find the derivative of f(x): $$f'(x) = u'(x)v(x)+u(x)v'(x)$$ Substituting the values we found in Steps 1 and 2, we get: $$f'(x) = (-2)e^{-x} + (1-2x)(-e^{-x})$$
04

Simplify the derivative

Finally, let's simplify the expression for f'(x): $$f'(x) = -2e^{-x} + e^{-x} - 2xe^{-x}$$ Factoring out the common term, \(e^{-x}\), we get: $$f'(x) = e^{-x}(-2+1-2x)$$ So the derivative of the function f(x) is: $$f'(x) = (1-2x)e^{-x} - 2e^{-x}$$

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Key Concepts

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

Derivative
In calculus, a derivative represents how a function changes as its input changes. It's a fundamental concept needed to understand rates of change, and knowing how to compute derivatives is crucial. For example, if you have a function like \(f(x) = (1-2x)e^{-x}\), finding its derivative tells us about the rate at which \(f(x)\) changes with \(x\). By employing basic differentiation rules, you can determine this rate effectively. Differentiation helps us understand the underlying behavior of the function, such as whether it is increasing, decreasing, or remaining constant within its domain.
Product Rule
The product rule is a technique used when differentiating functions that are the multiplication of two dependent functions. For instance, with the function \(f(x) = u(x)v(x)\), where \(u(x) = (1-2x)\) and \(v(x) = e^{-x}\), the product rule states that:
  • The derivative \(f'(x)\) is given by \(f'(x) = u'(x)v(x) + u(x)v'(x)\).
  • This basically involves taking the derivative of the first function and multiplying it by the second function, then adding the product of the first function and the derivative of the second function.
This rule helps break down complex functions into manageable parts, enabling easier differentiation.
Chain Rule
The chain rule is a vital tool in calculus for finding the derivative of composite functions. A composite function is one where a function is inside another function, like \(v(x) = e^{-x}\) in our example. To differentiate \(v(x)\), we need the chain rule because \(e^{-x}\) essentially is a function \(e^u\) where \(u = -x\). Here's how it works:
  • Differentiate the outer function, \(e^u\), with respect to \(u\), which gives us \(e^u\).
  • Then, differentiate the inner function \(-x\) with respect to \(x\), yielding \(-1\).
  • Finally, multiply these derivatives together to get \(-e^{-x}\).
Ultimately, the chain rule simplifies the process of tackling more complicated derivatives involving nested functions.
Differentiation
Differentiation is the process of finding a derivative and is pivotal in calculus. Whether you're dealing with simple polynomials or more complicated expressions involving products and compositions of functions, the concepts of differentiation help break down the problem. With our initial function \(f(x) = (1-2x)e^{-x}\), differentiation involves using rules like the product rule and the chain rule. By applying these rules correctly, we find that:
  • The derivative of \(u(x) = 1-2x\) is \(-2\).
  • The derivative of \(v(x) = e^{-x}\) is \(-e^{-x}\).
  • Combining these results with the product rule gives us the simplified expression for the derivative.
Differentiation is essential for exploring the unique features and behavior of functions, enabling us to compute areas, optimize problems, and understand motion.

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Most popular questions from this chapter

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals). A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=m x ; x^{2}+y^{2}=a^{2},\) where \(m\) and \(a\) are constants

Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\ln \frac{(2 x-1)(x+2)^{3}}{(1-4 x)^{2}}$$

The volume of a torus (doughnut or bagel) with an inner radius of \(a\) and an outer radius of \(b\) is \(V=\pi^{2}(b+a)(b-a)^{2} / 4\) a. Find \(d b / d a\) for a torus with a volume of \(64 \pi^{2}\). b. Evaluate this derivative when \(a=6\) and \(b=10\)

A store manager estimates that the demand for an energy drink decreases with increasing price according to the function \(d(p)=\frac{100}{p^{2}+1},\) which means that at price \(p\) (in dollars), \(d(p)\) units can be sold. The revenue generated at price \(p\) is \(R(p)=p \cdot d(p)\) (price multiplied by number of units). a. Find and graph the revenue function. b. Find and graph the marginal revenue \(R^{\prime}(p)\). c. From the graphs of \(R\) and \(R^{\prime}\), estimate the price that should be charged to maximize the revenue.

Calculate the derivative of the following functions (i) using the fact that \(b^{x}=e^{x \ln b}\) and (ii) by using logarithmic differentiation. Verify that both answers are the same. $$y=3^{x}$$
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