Chapter 3: Problem 53
\(45-54\) Find the derivative of the function. Simplify where possible. $$y=\cos ^{-1}\left(e^{2 x}\right)$$
Short Answer
Expert verified
\(\frac{dy}{dx} = -\frac{2e^{2x}}{\sqrt{1-e^{4x}}}\)
Step by step solution
01
Identify the Function
We are given the function \( y = \cos^{-1}(e^{2x}) \). This function involves the inverse cosine (arc cosine) and an exponential term \( e^{2x} \).
02
Apply the Chain Rule
To find the derivative, we need to employ the chain rule. The chain rule states:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\] where \( u = e^{2x} \).
03
Differentiate the Outside Function
The derivative of \( y = \cos^{-1}(u) \) with respect to \( u \) is \(-\frac{1}{\sqrt{1-u^2}}\). So, \( \frac{dy}{du} = -\frac{1}{\sqrt{1-u^2}} \).
04
Differentiate the Inside Function
Differentiate \( u = e^{2x} \) with respect to \( x \). The derivative \( \frac{du}{dx} \) is \( 2e^{2x} \) because the derivative of \( e^{kx} \) is \( ke^{kx} \).
05
Combine Using Chain Rule
Combine the expressions using the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{\sqrt{1-(e^{2x})^2}} \cdot 2e^{2x} \]
06
Simplification
Simplify the expression: \( (e^{2x})^2 = e^{4x} \). Thus, \[ \frac{dy}{dx} = -\frac{2e^{2x}}{\sqrt{1-e^{4x}}} \]. This is the derivative of the given function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
chain rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. Think of it as a way to peel back layers of an onion. It allows us to find the derivative where a function is nested inside another.
In this exercise, our nested function is \( y = \cos^{-1}(e^{2x}) \), where the inverse cosine function is applied to \( e^{2x} \).
To differentiate this composite function, we used the chain rule, which tells us that if we have \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) can be found by multiplying the derivative of the inside function by the derivative of the outside function:
In this exercise, our nested function is \( y = \cos^{-1}(e^{2x}) \), where the inverse cosine function is applied to \( e^{2x} \).
To differentiate this composite function, we used the chain rule, which tells us that if we have \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) can be found by multiplying the derivative of the inside function by the derivative of the outside function:
- First, identify the inner function \( u = e^{2x} \) and the outer function \( y = \cos^{-1}(u) \).
- Calculate the derivative of the outer function with respect to the inner function.
- Calculate the derivative of the inner function with respect to \( x \).
- Multiply these derivatives to find \( \frac{dy}{dx} \).
derivatives
Derivatives are the tools we use to find how a function is changing at any given point. They give us the slope of the tangent line to the graph of the function at a particular point, representing the rate of change.
When dealing with inverse trigonometric functions, their derivatives have specific forms. For instance, the derivative of \( \cos^{-1}(u) \) is \(-\frac{1}{\sqrt{1-u^2}}\).
This comes from understanding the arc cosine function. The negative sign denotes a decreasing function, due to the shape and behavior of \( \cos^{-1}(u) \).
When dealing with inverse trigonometric functions, their derivatives have specific forms. For instance, the derivative of \( \cos^{-1}(u) \) is \(-\frac{1}{\sqrt{1-u^2}}\).
This comes from understanding the arc cosine function. The negative sign denotes a decreasing function, due to the shape and behavior of \( \cos^{-1}(u) \).
- The derivative of a function can vary significantly based on its composition.
- Recognizing how different derivatives apply to different functions is crucial.
exponential functions
Exponential functions, like \( e^{kx} \), are powerful mathematical expressions featuring a constant base raised to a variable exponent. These functions have unique characteristics, such as continuous growth or decay, depending on the sign of the exponent.
In this scenario, we are dealing with \( e^{2x} \). The derivative of an exponential function \( e^{kx} \) is straightforward – it’s \( ke^{kx} \). The constant \( k \) multiplies the function itself, reflecting how rapidly the function's rate changes.
In this scenario, we are dealing with \( e^{2x} \). The derivative of an exponential function \( e^{kx} \) is straightforward – it’s \( ke^{kx} \). The constant \( k \) multiplies the function itself, reflecting how rapidly the function's rate changes.
- Combining exponential terms with other functions requires careful evaluating, especially in calculus.
- They often appear in various scientific models because of their natural growth representation.
simplification steps
Simplification is often the final touch needed to make a derivative or any mathematical expression more understandable and easier to work with. After deriving an expression, it’s crucial to simplify it.
For \( y = \cos^{-1}(e^{2x}) \), this involves combining multiple terms into a more manageable form. In our derivative:
For \( y = \cos^{-1}(e^{2x}) \), this involves combining multiple terms into a more manageable form. In our derivative:
- \((e^{2x})^2\) becomes \(e^{4x}\), simplifying the nested exponential part.
- The final expression \(-\frac{2e^{2x}}{\sqrt{1-e^{4x}}}\) is achieved by organizing terms into a standard fraction form, which is often clearer for interpretation.