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Find the derivative of the function. $$ f(x)=\sin ^{-1}\left(e^{2 x}\right) $$

Short Answer

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

Step by step solution

01

Identify the outer and inner functions

To find the derivative, first identify the outer function, which is the inverse sine function, and the inner function, which is the exponential function: Outer function: \( \sin^{-1}(u) \) Inner function: \( u = e^{2x} \)
02

Differentiate the outer function

Now, we will differentiate the outer function with respect to its argument: $$ \frac{d}{du}(\sin^{-1}(u)) = \frac{1}{\sqrt{1-u^2}} $$ Note that we used the derivative formula for the inverse sine function.
03

Differentiate the inner function

Next, differentiate the inner function with respect to x: $$ \frac{du}{dx} = \frac{d}{dx}\left(e^{2x}\right) = 2e^{2x} $$
04

Apply the Chain Rule

Now, we will apply the chain rule to find the derivative of the composite function: $$ \frac{df}{dx} = \frac{d}{dx}\left(\sin^{-1}\left(e^{2x}\right)\right) = \frac{d}{du}(\sin^{-1}(u)) \cdot \frac{du}{dx} $$
05

Substitute the derivatives and simplify

Substitute the derivatives of the outer and inner functions from Steps 2 and 3: $$ \frac{df}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot 2e^{2x} $$ Now, replace u with the inner function (\(u = e^{2x}\)): $$ \frac{df}{dx} = \frac{1}{\sqrt{1-e^{4x}}} \cdot 2e^{2x} $$ So, the derivative of the function f(x) 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.

Understanding the Chain Rule
The Chain Rule is a fundamental concept in calculus used to find the derivative of a composite function. A composite function is essentially a function within another function. For instance, if you have \( f(g(x)) \), it means the function \( g \) is embedded inside the function \( f \). Our task is to find how changing \( x \) affects the overall function \( f \).

To apply the Chain Rule, you first identify the functions involved. You'll have an 'outer' function and an 'inner' function. The outer function depends on the inner one.
  • The outer function is derived concerning its argument, giving \( \frac{df}{du} \).
  • The inner function is derived regarding \( x \), given by \( \frac{du}{dx} \).
By multiplying these derivatives, \( \frac{df}{du} \cdot \frac{du}{dx} \), you find the derivative of the composite function. This is the essence of the Chain Rule: break down a complex problem by separating it into smaller, manageable parts.
Deciphering Inverse Sine Function
The inverse sine function, often denoted as \( \sin^{-1}(x) \) or \( \arcsin(x) \), is the function that "undoes" what the sine function does. When you take the sine of an angle, you're mapping from an angle to a ratio. The inverse sine brings you back from a ratio to an angle.

The key to understanding its derivative is recognizing the limitation of its output values. Since the sine function's outputs are limited between -1 and 1, the inverse sine, \( \sin^{-1}(x) \), has a derivative that requires extra care to maintain the function's definitions. This derivative is: \[ \frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}. \]This formula helps determine the rate of change of the inverse sine function with respect to its argument.

The term inside the square root, \( 1-x^2 \), ensures the function remains valid only when \( x \) is within the interval \([-1,1]\), matching the range of sine. As you compute derivatives involving inverse sine, keeping this domain restriction in mind is crucial for accuracy.
Examining Exponential Functions
Exponential functions are crucial in mathematics, defined as functions where the variable appears in the exponent. Examples of exponential functions include \( e^x \) and its variations like \( e^{2x} \), where \( e \) is the base of the natural logarithm, approximately 2.718.

The power of exponential functions lies in their property of continuous growth or decay. For instance, if you consider the function \( e^{2x} \), here 2 is the rate at which the growth process occurs. Derivatives of exponential functions highlight their unique nature:
\[\frac{d}{dx}(e^{2x}) = 2e^{2x}.\]This formula reveals that the function's rate of change is proportional to its value, a characteristic enabling exponential functions to model real-world scenarios like population growth or radioactive decay.

When dealing with derivatives in the context of exponential functions, always remember: the derivative of \( e^{kx} \) is \( ke^{kx} \), where \( k \) is a constant. This straightforward characteristic makes working with exponential derivatives particularly enjoyable and predictable.

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Most popular questions from this chapter

A projectile is fired from a cannon located on a horizontal plane. If we think of the cannon as being located at the origin \(O\) of an \(x y\) -coordinate system, then the path of the projectile is $$y=\sqrt{3} x-\frac{x^{2}}{400}$$ where \(x\) and \(y\) are measured in feet. a. Find the value of \(\theta\) (the angle of elevation of the gun). b. At what point on the trajectory is the projectile traveling parallel to the ground? c. What is the maximum height attained by the projectile? d. What is the range of the projectile (the distance \(O A\) along the \(x\) -axis)? e. At what angle with respect to the \(x\) -axis does the projectile hit the ground?

Verify each differentiation formula. a. \(\frac{d}{d x} \cos ^{-1} u=-\frac{1}{\sqrt{1-u^{2}}} \frac{d u}{d x}\) b. \(\frac{d}{d x} \tan ^{-1} u=\frac{1}{1+u^{2}} \frac{d u}{d x}\) c. \(\frac{d}{d x} \csc ^{-1} u=-\frac{1}{|u| \sqrt{u^{2}-1}} \frac{d u}{d x}\) d. \(\frac{d}{d x} \sec ^{-1} u=\frac{1}{|u| \sqrt{u^{2}-1}} \frac{d u}{d x}\) e. \(\frac{d}{d x} \cot ^{-1} u=-\frac{1}{1+u^{2}} \frac{d u}{d x}\)

Acceleration of a Car A car starting from rest and traveling in a straight line attains a velocity of $$v(t)=\frac{110 t}{2 t+5}$$ feet per second after \(t\) sec. Find the initial acceleration of the car and its acceleration \(10 \mathrm{sec}\) after starting from rest.

Water flows from a tank of constant cross-sectional area \(50 \mathrm{ft}^{2}\) through an orifice of constant cross-sectional area \(\frac{1}{4} \mathrm{ft}^{2}\) located at the bottom of the tank. Initially, the height of the water in the tank was \(20 \mathrm{ft}\), and \(t\) sec later it was given by the equation $$ 2 \sqrt{h}+\frac{1}{25} t-2 \sqrt{20}=0 \quad 0 \leq t \leq 50 \sqrt{20} $$ How fast was the height of the water decreasing when its height was \(9 \mathrm{ft}\) ?

Find the derivative of the function. $$ y=\sec ^{-1} x+\csc ^{-1} x $$

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