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Calculate the derivative of the following functions. $$y=\left(1-e^{-0.05 x}\right)^{-1}$$

Short Answer

Expert verified
Answer: The derivative of the function is \(\frac{dy}{dx} = -0.05 e^{-0.05 x}\left(1-e^{-0.05x}\right)^{-2}\).

Step by step solution

01

Identify the outer and inner functions

In this problem, the outer function \(f(u) = u^{-1}\) and the inner function \(g(x) = 1 - e^{-0.05x}\). Now, we will find the derivatives of these two functions.
02

Find the derivative of the outer function

First, let's find the derivative of the outer function. For \(f(u) = u^{-1}\), according to the power rule, the derivative is: $$f'(u) = -u^{-2}$$
03

Find the derivative of the inner function

Now, let's find the derivative of the inner function. For \(g(x) = 1 - e^{-0.05x}\), the derivative can be found using the chain rule for \(-0.05x\) inside the exponential function: $$g'(x) = - (-0.05) e^{-0.05 x} = 0.05 e^{-0.05 x}$$
04

Apply the chain rule

Now that we have the derivatives of the outer and inner functions, we will apply the chain rule to find the derivative of the composite function: $$(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$$ Substitute the derivatives and functions into the chain rule: $$\frac{dy}{dx} = f'(1-e^{-0.05x}) \cdot 0.05 e^{-0.05 x}$$ Plug in \(f'(u) = -u^{-2}\) and simplify: $$\frac{dy}{dx} = -\left(1-e^{-0.05x}\right)^{-2} \cdot 0.05 e^{-0.05 x}$$
05

Write the final answer

The derivative of the given function is: $$\frac{dy}{dx} =-0.05 e^{-0.05 x}\left(1-e^{-0.05x}\right)^{-2}$$

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Key Concepts

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

Chain Rule
Understanding the chain rule is crucial in calculus, especially when dealing with composite functions, where one function is inside of another. It allows us to differentiate complex expressions by breaking them down into simpler ones.

When we have a composite function, we can think of it as a combination of two functions, say, an outer function, denoted by \(f\), and an inner function, denoted by \(g\). The derivative of the composite function \(f(g(x))\) is found by taking the derivative of the outer function with respect to the inner function and then multiplying it by the derivative of the inner function with respect to \(x\). In formula form, the chain rule states that \[\frac{d}{dx}f(g(x)) = f'(g(x)) \times g'(x)\].

To illustrate, consider the function \(y = (1-e^{-0.05x})^{-1}\). Here, the outer function is \(f(u) = u^{-1}\) and the inner function is \(g(x) = 1 - e^{-0.05x}\). According to the chain rule, we first find the derivatives of both functions separately, and then multiply them to get the derivative of the composite function. This process simplifies the differentiation task and is widely applicable in many areas where composite functions are present.
Exponential Functions
Exponential functions are a class of mathematical functions characterized by an expression of the form \(b^x\), where \(b\) is a positive constant base and \(x\) is any real number exponent. In the natural exponential function, the base is \(e\), which is an irrational number approximately equal to 2.71828, and holds a special place in calculus due to its unique properties in relation to the rate of change.

When it comes to finding the derivative of exponential functions, they have a remarkable trait: the derivative of an exponential function \(e^x\) is itself, \(e^x\). However, if the exponent is a function of \(x\), like \(e^{-0.05x}\), we must involve the chain rule. In the given function \(y = (1-e^{-0.05x})^{-1}\), the exponential part \(e^{-0.05x}\) is crucial. Its derivative is \(0.05 e^{-0.05x}\) due to the constant multiple rule applied in conjunction with the chain rule.

The exponential function's continuous growth or decay models many natural phenomena such as population growth, radioactive decay, and even financial investments, making its understanding important in both theory and real-world applications.
Power Rule
The power rule is a fundamental technique used to differentiate functions of the form \(x^n\), where \(n\) is any real number. This rule states that the derivative of \(x^n\) is \(nx^{n-1}\). It provides a quick and straightforward way to find the derivative of power functions without the need for limits.

Applying the power rule to the outer function \(f(u) = u^{-1}\) in the original exercise leads to the derivative \(f'(u) = -u^{-2}\), which is obtained by multiplying the exponent \(n=-1\) by the function and decreasing the exponent by one. The power rule, often used in conjunction with the chain rule, is an essential tool for calculating the derivatives of polynomial functions and, more broadly, any function where variables are raised to powers.

In practice, the power rule not only saves time but also simplifies complex calculations, making it easier for students and professionals alike to work efficiently with powers in differentiation.

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Most popular questions from this chapter

Find the following higher-order derivatives. $$\frac{d^{2}}{d x^{2}}\left(\log _{10} x\right)$$

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u. \end{array}\right.$$ a. Show that \(\lim _{x \rightarrow u} H(v)=0\) b. For any value of \(u\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u).$$ c. Show that. $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right).$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\).

Logistic growth Scientists often use the logistic growth function \(P(t)=\frac{P_{0} K}{P_{0}+\left(K-P_{0}\right) e^{-r_{d}}}\) to model population growth, where \(P_{0}\) is the initial population at time \(t=0, K\) is the carrying capacity, and \(r_{0}\) is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model. World population (part 1 ) The population of the world reached 6 billion in \(1999(t=0)\). Assume Earth's carrying capacity is 15 billion and the base growth rate is \(r_{0}=0.025\) per year. a. Write a logistic growth function for the world's population (in billions) and graph your equation on the interval \(0 \leq t \leq 200\) using a graphing utility. b. What will the population be in the year 2020? When will it reach 12 billion?

Rule In general, the derivative of a quotient is not the quotient of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f / g\) equals \(f^{\prime} / g^{\prime}\)

Suppose you own a fuel-efficient hybrid automobile with a monitor on the dashboard that displays the mileage and gas consumption. The number of miles you can drive with \(g\) gallons of gas remaining in the tank on a particular stretch of highway is given by \(m(g)=50 g-25.8 g^{2}+12.5 g^{3}-1.6 g^{4},\) for \(0 \leq g \leq 4\). a. Graph and interpret the mileage function. b. Graph and interpret the gas mileage \(m(g) / \mathrm{g}\). c. Graph and interpret \(d m / d g\).

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