Last updated on July 24th, 2025
The derivative of inverse hyperbolic functions is crucial in understanding how these functions change in response to variations in x. Derivatives are useful in various fields, from engineering to economics, where they help in optimizing solutions. We will now delve into the derivatives of inverse hyperbolic functions.
Inverse hyperbolic functions have well-defined derivatives that indicate their differentiability within their domains. The derivatives of these functions are essential in calculus and mathematical analysis. The key inverse hyperbolic functions and their derivatives are: Inverse Hyperbolic Sine: \( \sinh^{-1}(x) \) Inverse Hyperbolic Cosine: \( \cosh^{-1}(x) \) Inverse Hyperbolic Tangent: \( \tanh^{-1}(x) \)
The formulas for the derivatives of the inverse hyperbolic functions are: \( \frac{d}{dx}(\sinh^{-1}(x)) = \frac{1}{\sqrt{x^2 + 1}} \) \( \frac{d}{dx}(\cosh^{-1}(x)) = \frac{1}{\sqrt{x^2 - 1}} \) \( \frac{d}{dx}(\tanh^{-1}(x)) = \frac{1}{1 - x^2} \) These formulas are valid within their respective domains.
We can derive the derivatives of inverse hyperbolic functions using implicit differentiation and trigonometric identities. Here are some methods: By Implicit Differentiation Using Hyperbolic Identities We will demonstrate the differentiation of \( \sinh^{-1}(x) \) using implicit differentiation: By Implicit Differentiation Assume \( y = \sinh^{-1}(x) \), so \( x = \sinh(y) \). Differentiate both sides with respect to x: \( \frac{d}{dx}(x) = \cosh(y) \cdot \frac{dy}{dx} \) Since \( \cosh^2(y) - \sinh^2(y) = 1 \), \( \cosh(y) = \sqrt{x^2 + 1} \). \( 1 = \sqrt{x^2 + 1} \cdot \frac{dy}{dx} \) implies \( \frac{dy}{dx} = \frac{1}{\sqrt{x^2 + 1}} \). Hence, \( \frac{d}{dx}(\sinh^{-1}(x)) = \frac{1}{\sqrt{x^2 + 1}} \).
Higher-order derivatives provide insights into the behavior of inverse hyperbolic functions. The first derivative shows the rate of change, while the second derivative indicates concavity or convexity. Calculating higher-order derivatives involves repeated differentiation of the first derivative. For example, the second derivative of \( \sinh^{-1}(x) \) is obtained from the first derivative \( \frac{1}{\sqrt{x^2 + 1}} \).
For \( x = 0 \), the derivative of \( \sinh^{-1}(x) \) is \( \frac{1}{\sqrt{0^2 + 1}} = 1 \). For \( x = 1 \), the derivative of \( \tanh^{-1}(x) \) is undefined because it reaches an asymptote there.
Students often make errors when differentiating inverse hyperbolic functions. Here are some common mistakes and solutions:
Calculate the derivative of \( (\sinh^{-1}(x) \cdot \cosh^{-1}(x)) \).
Let \( f(x) = \sinh^{-1}(x) \cdot \cosh^{-1}(x) \). Using the product rule, \( f'(x) = u'v + uv' \). Here, \( u = \sinh^{-1}(x) \) and \( v = \cosh^{-1}(x) \). Differentiate each: \( u' = \frac{1}{\sqrt{x^2 + 1}} \), \( v' = \frac{1}{\sqrt{x^2 - 1}} \). Substituting, \( f'(x) = \left(\frac{1}{\sqrt{x^2 + 1}}\right) \cdot \cosh^{-1}(x) + \sinh^{-1}(x) \cdot \left(\frac{1}{\sqrt{x^2 - 1}}\right) \). Thus, \( f'(x) = \frac{\cosh^{-1}(x)}{\sqrt{x^2 + 1}} + \frac{\sinh^{-1}(x)}{\sqrt{x^2 - 1}} \).
We apply the product rule to differentiate the given function. By dividing it into two parts, we differentiate each part separately and combine them using the product rule.
A cable is suspended between two poles, forming a catenary described by \( y = \cosh^{-1}(x) \). If x = 2 meters, find the slope of the cable.
Given \( y = \cosh^{-1}(x) \), the slope is the derivative \( \frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}} \). Substitute \( x = 2 \): \( \frac{dy}{dx} = \frac{1}{\sqrt{2^2 - 1}} = \frac{1}{\sqrt{3}} \). The slope of the cable at x = 2 meters is \( \frac{1}{\sqrt{3}} \).
To find the slope of the cable, we differentiate the given catenary function and substitute \( x = 2 \) into the derivative to find the rate of change at that point.
Derive the second derivative of \( y = \tanh^{-1}(x) \).
First, find the first derivative: \( \frac{dy}{dx} = \frac{1}{1 - x^2} \). Now, differentiate again for the second derivative: \( \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{1 - x^2}\right) \). Using the chain rule, \( \frac{d^2y}{dx^2} = \frac{0 \cdot (1 - x^2) + 2x}{(1 - x^2)^2} = \frac{2x}{(1 - x^2)^2} \). Thus, the second derivative is \( \frac{2x}{(1 - x^2)^2} \).
We differentiate the first derivative using the chain rule to find the second derivative, carefully handling the expression involving \( x \).
Prove: \( \frac{d}{dx}(\cosh^{-1}(x^2)) = \frac{2x}{\sqrt{x^4 - 1}} \).
Start by applying the chain rule: Let \( y = \cosh^{-1}(x^2) \). Then, \( \frac{dy}{dx} = \frac{1}{\sqrt{x^4 - 1}} \cdot \frac{d}{dx}(x^2) \). \( \frac{d}{dx}(x^2) = 2x \). Substitute back: \( \frac{dy}{dx} = \frac{2x}{\sqrt{x^4 - 1}} \). Thus, proved.
We apply the chain rule to differentiate \( \cosh^{-1}(x^2) \), multiplying the derivative of the outer function by the derivative of the inner function.
Solve: \( \frac{d}{dx}(\sinh^{-1}(x)/x) \).
Use the quotient rule: \( \frac{d}{dx}\left(\frac{\sinh^{-1}(x)}{x}\right) = \frac{x \cdot \frac{1}{\sqrt{x^2 + 1}} - \sinh^{-1}(x) \cdot 1}{x^2} \). Simplify: \( = \frac{x}{x^2 \sqrt{x^2 + 1}} - \frac{\sinh^{-1}(x)}{x^2} \). Therefore, \( \frac{d}{dx}\left(\frac{\sinh^{-1}(x)}{x}\right) = \frac{x}{x^2 \sqrt{x^2 + 1}} - \frac{\sinh^{-1}(x)}{x^2} \).
The quotient rule is used to differentiate the given function. By simplifying the expression, we arrive at the final result.
Inverse Hyperbolic Functions: Functions that are inverses of hyperbolic functions, such as \( \sinh^{-1}(x) \), \( \cosh^{-1}(x) \), and \( \tanh^{-1}(x) \). Implicit Differentiation: A method to find derivatives by differentiating both sides of an equation with respect to a variable. Chain Rule: A rule for differentiating composite functions by multiplying the derivatives of the inner and outer functions. Hyperbolic Identities: Equations such as \( \cosh^2(y) - \sinh^2(y) = 1 \), used in differentiating hyperbolic functions. Quotient Rule: A formula to differentiate a quotient of two functions, expressed as \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \).
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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