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Chapter 3: Problem 50

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

Short Answer

Expert verified
Question: Determine the derivative of the function $$\left(1-e^{-0.05 x}\right)^{-1}$$ with respect to x. Answer: The derivative of the function is $$\frac{0.05 \cdot e^{-0.05x}}{\left(1-e^{-0.05 x}\right)^2}.$$

Step by step solution

01

Identify the outer and inner functions

First, let's recognize that this is a composite function, meaning there are two functions involved: one nested inside another. The outer function is \(f(u) = u^{-1}\), and the inner function is \(g(x) = 1 - e^{-0.05x}\). We'll use the chain rule to find the derivative with respect to x.
02

Find the derivative of the outer function

Using the power rule for differentiation, we'll find the derivative of the outer function with respect to u: $$f'(u) = -u^{-2} = -\frac{1}{u^2}$$
03

Find the derivative of the inner function

Now, we'll find the derivative of the inner function with respect to x, using the chain rule again: $$g'(x) = 0 - e^{-0.05x} \cdot(-0.05) = 0.05 \cdot e^{-0.05x}$$
04

Apply the chain rule

Finally, we can apply the chain rule, multiplying the derivatives of the inner and the outer functions: $$\frac{d}{dx}\left[\left(1-e^{-0.05 x}\right)^{-1}\right] = f'(g(x)) \cdot g'(x)$$ Substitute \(f'(u)\) and \(g'(x)\): $$\frac{d}{dx}\left[\left(1-e^{-0.05 x}\right)^{-1}\right] = -\frac{1}{(1-e^{-0.05 x})^2} \cdot 0.05 \cdot e^{-0.05x}$$
05

Simplify the result

Multipling the two expressions, we arrive at the final simplified form of the derivative: $$\frac{d}{dx}\left[\left(1-e^{-0.05 x}\right)^{-1}\right] = \frac{0.05 \cdot e^{-0.05x}}{\left(1-e^{-0.05 x}\right)^2}$$

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Key Concepts

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

Composite Functions
Composite functions are formed when one function is nested inside another. In the problem, the function to differentiate is \[\left(1-e^{-0.05 x}\right)^{-1}\].Here, the outer and inner functions need to be identified. The outer function is \(f(u) = u^{-1}\),which takes a result from the inner function \(g(x) = 1 - e^{-0.05x})\).
Composite functions are essential in calculus when expressions seem complicated due to multiple layers of functions.
  • Outer functions are those that operate on the result of inner functions.
  • Inner functions provide the inputs for outer functions.
The beauty of composite functions lies in their versatility across various mathematics and real-world scenarios.
Power Rule
The power rule is a basic guideline for finding the derivative of functions of the form \(x^n\).This rule states that if you have a function \(f(x) = x^n\),then the derivative \(f'(x)\) is calculated as \(nx^{n-1}\).
For the function \(f(u) = u^{-1}\),the power rule helps to find its derivative.Applying the power rule, we change the exponent from \(-1\) to \(-2\) and multiply by the original exponent, giving us:\[f'(u) = -u^{-2} = -\frac{1}{u^2}\].Remember, the power rule simplifies many differentiation problems, making it an indispensable tool in calculus.
Derivative
Derivatives are a core concept in calculus used to measure how a function changes as its input changes. The derivative of a function at a particular point provides the slope of the tangent line to the function at that point.
The problem involves calculating the derivative of a composite function using the derivatives of both the outer and inner functions.
  • Differentiate the outer function using the power rule.
  • Differentiate the inner function separately.
The final step involves applying the chain rule, a fundamental rule dealing with composite functions. The derivative expresses the rate of change or the sensitivity of one quantity to change in another, making it tremendously useful in numerous fields.
Differential Calculus
Differential calculus focuses on how functions change and involves the study of rates and slopes. The chain rule, used in the solution, is a key component of differential calculus. It allows us to differentiate composite functions systematically.
By combining \(f'(u)\) and \(g'(x)\)from our problem:\[\frac{d}{dx}\left[\left(1-e^{-0.05 x}\right)^{-1}\right] = -\frac{1}{(1-e^{-0.05 x})^2} \cdot 0.05 \cdot e^{-0.05x}\],we demonstrate differential calculus in action.
  • The focus is on real-world scenarios where change needs quantifying.
  • Differential calculus provides tools to work with continuously varying quantities.
The final derivative reveals the relationship between variables, showcasing differential calculus' power to dissect and understand complex problems.

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Most popular questions from this chapter

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