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In Exercises 15-28, find the derivative of the function. $$ g(x)=e^{x} \arcsin x $$

Short Answer

Expert verified
The derivative of the function \(g(x)=e^{x} \arcsin x\) is \((e^{x} *\arcsin x)' = e^{x}*\frac{1}{\sqrt{1-x^{2}}} + e^{x}*\arcsin x\).

Step by step solution

01

Identify the Components of the Function

Recognize that the function \(g(x)=e^{x} \arcsin x\) is a product of two separate functions: \(f(x)=e^{x}\) and \(g(x)=\arcsin x\). These will be differentiated separately, using the product rule and the chain rule.
02

Differentiate the exponential function

The derivative of \(f(x)=e^{x}\) is \(f'(x)=e^{x}\).
03

Differentiate the arcsine function

To find the derivative of the arcsin function, we’ll use the following formula for the derivative of arcsin: \((\arcsin u)' = \frac{u'}{\sqrt{1-u^{2}}}\). Here \(u=x\), so \(u'=1\). Therefore, \((\arcsin x)' = \frac{1}{\sqrt{1-x^{2}}}\). So, the derivative of \(g(x)=\arcsin x\) is \(g'(x)=\frac{1}{\sqrt{1-x^{2}}}\).
04

Apply the Product Rule

Use the formula for the derivative of a product of two functions: \((f*g)' = f'g + f*g'\). Substituting the functions and their derivatives, we get \((e^{x} *\arcsin x)' = e^{x}*\frac{1}{\sqrt{1-x^{2}}} + e^{x}*\arcsin x\).

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Key Concepts

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

Product Rule
When tasked with finding the derivative of a function that is a product of two separate functions, such as g(x) = e^x \cdot \arcsin x, we must make use of the product rule. This rule is a crucial part of calculus and it tells us that the derivative of the product f(x) \cdot g(x) is not simply the product of the derivatives of f(x) and g(x). Instead, the product rule states:

\[\begin{equation}(f \cdot g)' = f' \cdot g + f \cdot g'\end{equation}\]
This means we need to first take the derivative of the first function (f(x)) and multiply it by the second function (g(x)) as it is, then we add the product of the first function as it is with the derivative of the second function (g'(x)). Applying the product rule correctly is essential for accurately calculating the derivative of a product of functions.
Chain Rule
The chain rule is another fundamental concept in calculus, particularly when dealing with composite functions. A composite function is made up when one function is applied within another, and in the case of differentiation, we cannot simply differentiate the outer function without considering the inner function. The chain rule tells us how to take the derivative of a composite function. It states that if you have a composite function h(x) = f(g(x)), then the derivative of h(x) with respect to x is:

\[\begin{equation} h'(x) = f'(g(x)) \cdot g'(x)\end{equation}\]
In other words, it's the derivative of the outer function evaluated at the inner function, times the derivative of the inner function. When we differentiate the arcsine function or other trigonometric functions, the chain rule often comes into play when these functions are expressed in terms of another function.
Derivative of Exponential Function
Exponential functions are common in mathematics, particularly those with base e, the mathematical constant approximately equal to 2.71828. The beauty of the natural exponential function f(x)=e^x lies in its simplicity when differentiating. The derivative of e^x with respect to x is:

\[\begin{equation}f'(x) = e^x\end{equation}\]
This characteristic makes it an ideal and straightforward case when applying the product or chain rule, as there's no additional manipulation required unlike other functions. The exponential function does not change when differentiated, which is unique compared to most other types of functions.
Derivative of Arcsin Function
The arcsine function represents the inverse of the sine function and is often represented as \arcsin x or \sin^{-1} x. It's not as straightforward to differentiate as the exponential function. Whenever we need to find the derivative of the arcsin function, we use a specific formula:

\[\begin{equation}(\arcsin u)' = \frac{u'}{\sqrt{1-u^2}}\end{equation}\]
In the case where u = x and u' = 1, this simplifies to:

\[\begin{equation}(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}\end{equation}\]
Understanding this derivative is important especially in cases where arcsin is part of a product of functions or a chain of functions. It's also essential to note that the domain of x for the arcsine function is limited to [-1, 1], since it represents angles whose sine gives x.

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Most popular questions from this chapter

A 15 -centimeter pendulum moves according to the equation \(\theta=0.2 \cos 8 t,\) where \(\theta\) is the angular displacement from the vertical in radians and \(t\) is the time in seconds. Determine the maximum angular displacement and the rate of change of \(\theta\) when \(t=3\) seconds.

Existence of an Inverse Determine the values of \(k\) such that the function \(f(x)=k x+\sin x\) has an inverse function.

The normal daily maximum temperatures \(T\) (in degrees Fahrenheit) for Denver, Colorado, are shown in the table. (Source: National Oceanic and Atmospheric Administration). $$ \begin{aligned} &\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Month } & \text { Jan } & \text { Feb } & \text { Mar } & \text { Apr } & \text { May } & \text { Jun } \\ \hline \text { Temperature } & 43.2 & 47.2 & 53.7 & 60.9 & 70.5 & 82.1 \\ \hline \end{array}\\\ &\begin{array}{|l|c|c|c|c|c|c|} \hline \text { Month } & \text { Jul } & \text { Aug } & \text { Sep } & \text { Oct } & \text { Nov } & \text { Dec } \\ \hline \text { Temperature } & 88.0 & 86.0 & 77.4 & 66.0 & 51.5 & 44.1 \\ \hline \end{array} \end{aligned} $$(a) Use a graphing utility to plot the data and find a model for the data of the form \(T(t)=a+b \sin (\pi t / 6-c)\) where \(T\) is the temperature and \(t\) is the time in months, with \(t=1\) corresponding to January. (b) Use a graphing utility to graph the model. How well does the model fit the data? (c) Find \(T^{\prime}\) and use a graphing utility to graph the derivative. (d) Based on the graph of the derivative, during what times does the temperature change most rapidly? Most slowly? Do your answers agree with your observations of the temperature changes? Explain.

In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{f(x)=\frac{1}{3} x \sqrt{x^{2}+5}} \quad \frac{\text { Point }}{(2,2)}\)

Linear and Quadratic Approximations The linear and quadratic approximations of a function \(f\) at \(x=a\) are \(P_{1}(x)=f^{\prime}(a)(x-a)+f(a)\) and \(P_{2}(x)=\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}+f^{\prime}(a)(x-a)+f(a)\) \(\begin{array}{llll}\text { In Exercises } & 133-136, & \text { (a) find the specified linear and }\end{array}\) quadratic approximations of \(f,\) (b) use a graphing utility to graph \(f\) and the approximations, (c) determine whether \(P_{1}\) or \(P_{2}\) is the better approximation, and (d) state how the accuracy changes as you move farther from \(x=a\). $$ \begin{array}{l} f(x)=x \ln x \\ a=1 \end{array} $$

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