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Chapter 3: Problem 58

Calculate the derivative of the following functions. $$y=\log _{8}|\tan x|$$

Short Answer

Expert verified
Question: Determine the derivative of the function, $$y = \frac{1}{1-e^{-0.05x}}$$, with respect to x. Answer: The derivative of the function with respect to x is $$\frac{dy}{dx} = \frac{0.05e^{-0.05x}}{(1-e^{-0.05x})^2}$$.

Step by step solution

01

Recognize the chain rule structure

We can see that the function has a composition of functions in the form of $$y = [f(g(x))]$$, where $$f(x) = (1-x)^{-1}$$ and $$g(x) = e^{-0.05x}$$. We will use the chain rule, which is $$\frac{dy}{dx} = \frac{df}{dg}\cdot \frac{dg}{dx}$$.
02

Calculate the derivative of f with respect to g

We will find the derivative of $$f(g) = (1-g)^{-1}$$ with respect to g. Using the exponent rule, we get $$\frac{df}{dg} = -1(1-g)^{-1-1} = -(1-g)^{-2}$$
03

Calculate the derivative of g with respect to x

We will find the derivative of $$g(x) = e^{-0.05x}$$ with respect to x. Using the exponent rule, we get $$\frac{dg}{dx} = -0.05e^{-0.05x}$$
04

Multiply the results from step 2 and 3 using the chain rule

Now we will use the chain rule to find the derivative of the function. We multiply the results from step 2 and 3: $$\frac{dy}{dx} = \frac{df}{dg}\cdot \frac{dg}{dx} = -(1-g)^{-2}\cdot(-0.05e^{-0.05x})$$
05

Substitute g(x) back into the equation and simplify

Finally, we substitute $$g(x) = e^{-0.05x}$$ back into the equation: $$\frac{dy}{dx} = -(1-e^{-0.05x})^{-2}\cdot(-0.05e^{-0.05x})$$ Now, we can simplify the expression: $$\frac{dy}{dx} = \frac{0.05e^{-0.05x}}{(1-e^{-0.05x})^2}$$ Now we have found the derivative of the given function: $$\frac{dy}{dx} = \frac{0.05e^{-0.05x}}{(1-e^{-0.05x})^2}$$

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Most popular questions from this chapter

Gravitational force The magnitude of the gravitational force between two objects of mass \(M\) and \(m\) is given by \(F(x)=-\frac{G M m}{x^{2}},\) where \(x\) is the distance between the centers of mass of the objects and \(G=6.7 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant (N stands for newton, the unit of force; the negative sign indicates an attractive force). a. Find the instantaneous rate of change of the force with respect to the distance between the objects. b. For two identical objects of mass \(M=m=0.1 \mathrm{kg},\) what is the instantaneous rate of change of the force at a separation of \(x=0.01 \mathrm{m} ?\) c. Does the instantaneous rate of change of the force increase or decrease with the separation? Explain.

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