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Evaluate the derivatives of the following functions. $$f(x)=\sin ^{-1}\left(e^{-2 x}\right)$$

Short Answer

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Question: Find the derivative of the function $$f(x)=\sin ^{-1}\left(e^{-2 x}\right)$$. Answer: The derivative of the function is $$f'(x) = \frac{-2e^{-2x}}{\sqrt{1 - e^{-4x}}}$$.

Step by step solution

01

Differentiating g(x) and f(x)

First, let's find the derivatives of $$g(x) = \sin^{-1}(x)$$ and $$f(x) = e^{-2x}$$ separately. For $$g(x) = \sin^{-1}(x)$$, we have: $$g'(x) = \frac{1}{\sqrt{1 - x^2}}$$ For $$f(x) = e^{-2x}$$, we have: $$f'(x) = -2e^{-2x}$$
02

Applying the Chain Rule

Now that we have the derivatives of $$g(x)$$ and $$f(x)$$, we can apply the chain rule. Recall that our composite function is $$h(x) = g(f(x))$$, so we want to find $$h'(x) = g'(f(x)) \cdot f'(x)$$. Using the derivatives found in Step 1: $$h'(x) = \frac{1}{\sqrt{1 - (e^{-2x})^2}} \cdot (-2e^{-2x})$$
03

Simplifying the Result

Now we will simplify the expression for the derivative: $$h'(x) = \frac{-2e^{-2x}}{\sqrt{1 - e^{-4x}}}$$ So, the derivative of the given function $$f(x)=\sin ^{-1}\left(e^{-2 x}\right)$$ is: $$f'(x) = \frac{-2e^{-2x}}{\sqrt{1 - e^{-4x}}}$$

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Key Concepts

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

Inverse Trigonometric Functions
Inverse trigonometric functions are special mathematical functions that allow us to determine angles when we know the values of trigonometric ratios. Unlike regular trigonometric functions like sine or cosine, these functions provide the angle itself as the output. For example, if you have the value of the sine and you want to find the angle that produces this sine value, you would use the inverse sine function, often denoted as \( \sin^{-1}(x) \).
In the original problem, the function to be differentiated is \( \sin^{-1}(e^{-2x}) \). Here, the outer function is the inverse sine and the inner function is an exponential function. Knowing how to differentiate inverse trigonometric functions is crucial because they appear in various mathematical and engineering fields.
These inverse functions have specific derivatives which are essential to remember. For \( \sin^{-1}(x) \), the derivative is \( \frac{1}{\sqrt{1-x^2}} \). This is a key formula in differentiation, which allows us to handle inverse trigonometric functions systematically.
Chain Rule
The chain rule is an essential differentiation technique used when you have a function nested inside another function, which is known as a composite function. It allows you to find the derivative of such functions efficiently. The rule states that if you have a function \( y = g(f(x)) \), then the derivative \( y' \) can be found using \( y' = g'(f(x)) \cdot f'(x) \).
In this problem, our function is \( f(x) = \sin^{-1}(e^{-2x}) \). Here, the inverse sine function \( g(x) = \sin^{-1}(x) \) is the outer function, and the exponential \( f(x) = e^{-2x} \) is the inner function. By the chain rule, we first differentiate \( g(x) \) as \( g'(x) = \frac{1}{\sqrt{1-x^2}} \) and \( f(x) \) as \( f'(x) = -2e^{-2x} \).
Then, we substitute \( e^{-2x} \) into the derivative of \( g \) and multiply it by \( f' \) to get the final derivative of the composite function. The chain rule is powerful because it simplifies the process of differentiating complex expressions involving nested functions.
Differentiation Techniques
Differentiation techniques are tools and methods we use to calculate the derivative of different functions. They form the foundation for understanding how rapidly a function changes at any point. Knowing various differentiation techniques helps solve a wide variety of problems in calculus.
Some basic techniques include the power rule, product rule, quotient rule, and chain rule. Each serves specific types of functions, and learning when to use each technique is key. For example, the chain rule is used for composite functions, as seen in the inverse trigonometric example, while the product rule is for differentiating products of functions.
In the original solution, you see a combination of differentiation techniques. The inverse trigonometric function derivative is directly applied with its standard formula, and the exponential function is differentiated using the basic rules for exponential differentiation. Then, the chain rule ties them together neatly, allowing you to find the overall derivative effectively.
  • This involves identifying the type of functions and choosing the right technique.
  • Practicing these techniques builds intuition, which is useful for handling complex problems.
  • Understanding each method's purpose helps avoid mistakes in calculations.
Mastering these techniques significantly aids in tackling a wide range of calculus problems.

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Most popular questions from this chapter

Identifying functions from an equation. The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots\) c. Use the functions found in part (b) to graph the given equation. $$x^{4}=2\left(x^{2}-y^{2}\right) \text { (eight curve) }$$

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

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?

\(F=f / g\) be the quotient of two functions that are differentiable at \(x\) a. Use the definition of \(F^{\prime}\) to show that $$ \frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{h g(x+h) g(x)} $$ b. Now add \(-f(x) g(x)+f(x) g(x)\) (which equals 0 ) to the numerator in the preceding limit to obtain $$\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}$$ Use this limit to obtain the Quotient Rule. c. Explain why \(F^{\prime}=(f / g)^{\prime}\) exists, whenever \(g(x) \neq 0\)

Find the following higher-order derivatives. $$\left.\frac{d^{3}}{d x^{3}}\left(x^{4.2}\right)\right|_{x=1}$$

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