Chapter 4: Problem 56
Find the derivative of the function.
\(y=\operatorname{sech}^{-1}(\cos 2 x), \quad 0
Short Answer
Expert verified
\( y' = \frac{2\sin(2x)}{\sqrt{1-\cos^2(2x)}|\cos(2x)|} \)
Step by step solution
01
Identify the outer and inner functions
Identify the 'outer' and 'inner' functions in the expression. The outer function is \( \operatorname{sech}^{-1}(u) \) and the inner function is \( \cos(2x) \). The derivative can be found by differentiating the outer function and multiplying it with the derivative of the inner function.
02
Differentiate the outer function
The derivative of a function \( \operatorname{sech}^{-1}(u) \) is \( -\frac{1}{|u|\sqrt{u^2-1}} \), and since here \( u = \cos(2x) \), derivative becomes \( -\frac{1}{|\cos(2x)|\sqrt{\cos^2(2x)-1}} \).
03
Differentiate the inner function
The inner function is \( \cos(2x) \). The derivative of \( \cos(2x) \) is \( -2\sin(2x) \).
04
Apply Chain rule
Multiply the derivative of the outer function by the derivative of the inner function to obtain the derivative of the composite function. Therefore, \( y' = \frac{2\sin(2x)}{\sqrt{1-\cos^2(2x)}|\cos(2x)|} \).
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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 tool in calculus for taking the derivative of composite functions. When one function is nested within another, the chain rule allows us to differentiate the entire expression by taking the derivative of the outer function and multiplying it by the derivative of the inner function.
To apply the chain rule, you first need to identify the 'outer' and 'inner' components of the function. For instance, if we have a function like \( f(g(x)) \), \( f \) would be the outer function, and \( g \) would be the inner function. The derivative of \( f(g(x)) \), denoted by \( \frac{d}{dx}[f(g(x))] \), is then \( f'(g(x)) \cdot g'(x) \), where \( f'(g(x)) \) is the derivative of the outer function evaluated at \( g(x) \), and \( g'(x) \) is the derivative of the inner function with respect to \( x \).
This technique is particularly useful when dealing with inverse hyperbolic functions, as these can often involve nested functions. For example, in the function \( y=\operatorname{sech}^{-1}(\cos 2x) \), the chain rule allows us to find the derivative efficiently by differentiating the inverse hyperbolic secant function and multiplying by the derivative of the cosine function.
To apply the chain rule, you first need to identify the 'outer' and 'inner' components of the function. For instance, if we have a function like \( f(g(x)) \), \( f \) would be the outer function, and \( g \) would be the inner function. The derivative of \( f(g(x)) \), denoted by \( \frac{d}{dx}[f(g(x))] \), is then \( f'(g(x)) \cdot g'(x) \), where \( f'(g(x)) \) is the derivative of the outer function evaluated at \( g(x) \), and \( g'(x) \) is the derivative of the inner function with respect to \( x \).
This technique is particularly useful when dealing with inverse hyperbolic functions, as these can often involve nested functions. For example, in the function \( y=\operatorname{sech}^{-1}(\cos 2x) \), the chain rule allows us to find the derivative efficiently by differentiating the inverse hyperbolic secant function and multiplying by the derivative of the cosine function.
Hyperbolic Functions
Hyperbolic functions are analogs of the ordinary trigonometric, or circular, functions, but instead of being defined with a circle, they are defined in terms of the hyperbola.
The most common hyperbolic functions include hyperbolic sine (\( \sinh(x) \)), hyperbolic cosine (\( \cosh(x) \)), and hyperbolic tangent (\( \tanh(x) \)), among others. They are used to describe the shapes of natural phenomena such as the distribution of electric potential between two opposite charges and the shape of a hanging cable (catenary).
The inverse hyperbolic functions, like the inverse hyperbolic secant (\( \operatorname{sech}^{-1}(x) \)), appear less frequently but play a significant role in solving certain differential equations and integrals. In calculating derivatives for inverse hyperbolic functions, it's important to remember their specific derivative formulas. For instance, the derivative of the inverse hyperbolic secant as used in the exercise is a combination of absolute value and square root expressions as seen in the solution step.
The most common hyperbolic functions include hyperbolic sine (\( \sinh(x) \)), hyperbolic cosine (\( \cosh(x) \)), and hyperbolic tangent (\( \tanh(x) \)), among others. They are used to describe the shapes of natural phenomena such as the distribution of electric potential between two opposite charges and the shape of a hanging cable (catenary).
The inverse hyperbolic functions, like the inverse hyperbolic secant (\( \operatorname{sech}^{-1}(x) \)), appear less frequently but play a significant role in solving certain differential equations and integrals. In calculating derivatives for inverse hyperbolic functions, it's important to remember their specific derivative formulas. For instance, the derivative of the inverse hyperbolic secant as used in the exercise is a combination of absolute value and square root expressions as seen in the solution step.
Trigonometric Derivatives
Trigonometric functions are foundational in mathematics, with applications spanning from geometry to physics. In calculus, understanding the derivatives of these functions is essential for analyzing periodic motions and waves.
Each trigonometric function, such as sine (\( \sin(x) \)), cosine (\( \cos(x) \)), and tangent (\( \tan(x) \)), has a corresponding derivative. For example, the derivative of \( \sin(x) \) is \( \cos(x) \), and the derivative of \( \cos(x) \) with respect to \( x \) is \( -\sin(x) \).
When these functions are involved in composite operations, as in the given exercise with \( \cos(2x) \), we must apply the chain rule while also considering their trigonometric derivatives. The differentiation of \( \cos(2x) \) would involve not only using the derivative of cosine, which is negative sine, but also taking into account the chain rule's impact due to the inner function \( 2x \), resulting in a final derivative of \( -2\sin(2x) \).
Each trigonometric function, such as sine (\( \sin(x) \)), cosine (\( \cos(x) \)), and tangent (\( \tan(x) \)), has a corresponding derivative. For example, the derivative of \( \sin(x) \) is \( \cos(x) \), and the derivative of \( \cos(x) \) with respect to \( x \) is \( -\sin(x) \).
When these functions are involved in composite operations, as in the given exercise with \( \cos(2x) \), we must apply the chain rule while also considering their trigonometric derivatives. The differentiation of \( \cos(2x) \) would involve not only using the derivative of cosine, which is negative sine, but also taking into account the chain rule's impact due to the inner function \( 2x \), resulting in a final derivative of \( -2\sin(2x) \).
