Question

For the following exercises, find the derivatives tor the functions.\(\sinh ^{-1}(\cosh (x))\)

Short answer

The derivative of \( \sinh^{-1}(\cosh(x)) \) is \( \tanh(x) \).

Step-by-step solution

Identify Inner and Outer Functions

The given function is \( \sinh^{-1}(\cosh(x)) \). This involves an inverse hyperbolic sine (\( \sinh^{-1} \)) applied to a hyperbolic cosine function (\( \cosh \)). Recognize that the inner function is \( \cosh(x) \) and the outer function is \( \sinh^{-1}(u) \), where \( u = \cosh(x) \).

Differentiate Outer Function

To differentiate \( \sinh^{-1}(u) \) with respect to \( u \), we use the derivative formula: \( \frac{d}{du} \sinh^{-1}(u) = \frac{1}{\sqrt{u^2 + 1}} \). In our problem, replace \( u \) with \( \cosh(x) \).

Chain Rule Application

Apply the chain rule to find \( \frac{d}{dx} \sinh^{-1}(\cosh(x)) \). According to the chain rule:\[ \frac{d}{dx} \sinh^{-1}(\cosh(x)) = \frac{d}{du} \sinh^{-1}(u) \cdot \frac{d}{dx} \cosh(x) \]where \( \frac{d}{dx} \cosh(x) = \sinh(x) \).

Substitute and Simplify

Substitute the derivatives calculated into the chain rule expression:\[ \frac{d}{dx} \sinh^{-1}(\cosh(x)) = \frac{1}{\sqrt{(\cosh(x))^2 + 1}} \cdot \sinh(x) \]Simplifying, we know \( (\cosh(x))^2 - 1 = \sinh^2(x) \), thus we have:\[ \frac{d}{dx} \sinh^{-1}(\cosh(x)) = \frac{\sinh(x)}{\sqrt{\sinh^2(x) + 1}} \]Since the square root approximates \( \cosh(x) \), it further simplifies the expression.

Final Derivative Expression

By simplifying further and using the identity \( \sinh^2(x) + 1 = \cosh^2(x) \), the derivative reduces to \( \frac{\sinh(x)}{\cosh(x)} = \tanh(x) \). Thus, the final derivative of \( \sinh^{-1}(\cosh(x)) \) is \( \tanh(x) \).

Key concepts

Derivative

In calculus, the term derivative represents the rate at which a function is changing at any given point. For a function \( f(x) \), the derivative, denoted \( f'(x) \) or \( \frac{df}{dx} \), represents the slope of the tangent line to the graph of the function at point \( x \).
This concept is foundational in calculus as it measures how a small change in one variable affects another.

• If \( f(x) \) is increasing at \( x \), \( f'(x) > 0 \).
• If \( f(x) \) is decreasing, \( f'(x) < 0 \).
• Where \( f(x) \) remains constant, \( f'(x) = 0 \).

In our exercise, finding the derivative involves identifying the rate of change of \( \sinh^{-1}(\cosh(x)) \) concerning \( x \). This requires understanding not just the function itself, but also the components involved like hyperbolic and inverse hyperbolic functions.

Hyperbolic Functions

Hyperbolic functions, akin to trigonometric functions, play a crucial role in calculus and complex analysis. They include \( \sinh(x) \) and \( \cosh(x) \), similar in form to the sine and cosine functions but based on exponential equations.

• The hyperbolic sine, \( \sinh(x) = \frac{e^x - e^{-x}}{2} \) gives a wave-like curve in hyperbolic geometry.
• The hyperbolic cosine, \( \cosh(x) = \frac{e^x + e^{-x}}{2} \) creates the shape of a catenary in physics and engineering.

These functions come with unique identities and properties, such as \( \cosh^2(x) - \sinh^2(x) = 1 \), paralleling the trigonometric identity \( \cos^2(x) + \sin^2(x) = 1 \). In the problem at hand, \( \cosh(x) \) is the inner function, crucial for applying the chain rule and simplifying the derivative.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are the inverses of the hyperbolic functions, such as \( \sinh^{-1}(x) \), often used to solve equations involving initially undetermined values.Their notations can also include \( \text{arsinh}(x) \), \( \text{arcosh}(x) \), etc. They help reverse hyperbolic transformations, akin to inverse trigonometric functions.For \( \sinh^{-1}(x) \), recognizing its derivative, \( \frac{d}{dx} \sinh^{-1}(x) = \frac{1}{\sqrt{x^2 + 1}} \), forms a key step in solving calculus problems.In our exercise, using the derivative of \( \sinh^{-1} \) with the chain rule helps us efficiently differentiate \( \sinh^{-1}(\cosh(x)) \).
This highlights its role in converting complex processes into simpler, derivable forms.

Chain Rule

The chain rule is a fundamental technique in calculus, allowing us to differentiate compositions of functions. It states that if a function \( y = f(g(x)) \) is the composition of \( f \) and \( g \), the derivative \( \frac{dy}{dx} \) is given by:\[ \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \]This rule helps break down complex functions into simpler parts, making them manageable in terms of differentiation.In the provided exercise, correctly applying the chain rule involves:

• Recognizing the outer function \( \sinh^{-1}(u) \) and the inner function \( \cosh(x) \)
• Computing the derivative of each
• Combining these to achieve the derivative \( \frac{d}{dx} \sinh^{-1}(\cosh(x)) = \tanh(x) \)

This streamlined process offers a methodical approach to tackling derivative problems involving multiple layers of functions.