Question

In Exercises \(13-22,\) find the derivatives of \(y\) with respect to the appropriate variable. $$y=\tan ^{-1} \sqrt{x^{2}-1}+\csc ^{-1} x, \quad x>1$$

Short answer

The derivative of the function \[ y=\tan ^{-1} \sqrt{x^{2}-1}+\csc ^{-1} x, \quad x>1 \] is \( \frac{dy}{dx} = \frac{x}{(x^{2}-1)^{3/2}} - \frac{1}{\sqrt{x^2-1}} \).

Step-by-step solution

Identify the Functions

First, identify the functions involved. The given function \(y\) consists of two terms: \(y=\tan ^{-1} \sqrt{x^{2}-1}\) and \(y= \csc ^{-1} x\). We will find their derivatives separately.

Differentiation of the first term

Differentiate the first term using the formula for the derivative of the inverse tangent, \( \frac{dy}{dx}= \frac{1}{1+u^2} \frac{du}{dx} \). Here \( u = \sqrt{x^{2}-1} \). Differentiating \( u \) with respect to \( x \) gives \( u' = \frac{x}{\sqrt{x^{2}-1}} \). Therefore, \( \frac{dy}{dx} = \frac{x}{\sqrt{x^{2}-1}(1 + x^{2} - 1)} \). Simplifying this gives \( \frac{dy}{dx} = \frac{x}{(x^{2}-1)^{3/2}} \).

Differentiation of the second term

Differentiate the second term using the formula for the derivative of the inverse cosecant, \( \frac{dy}{dx}= - \frac{1}{\sqrt{x^2-1}} \). Here the variable is \( x \), so no chain rule is necessary.

Combine the Results

Greater than one \( x \) requires \( x^2 -1 > 0 \). As a result, the derivative of the function is equal to the sum of the derivatives of each term: \( \frac{dy}{dx} = \frac{x}{(x^{2}-1)^{3/2}} - \frac{1}{\sqrt{x^2-1}} \).

Key concepts

Differentiation Techniques

Understanding differentiation techniques is crucial in calculus as it allows us to find the rate of change or slope of a function at any point. In our exercise, two techniques are central: the use of the derivative formulas for inverse trigonometric functions and the application of these formulas within the context of the given functions. Here, identifying the inner function for each term is essential. For the first term, involving inverse tangent, the inner function is the square root function, leading us to apply the chain rule after using the formula for the derivative of the inverse tangent. For the second term, the inverse cosecant, the derivative formula directly applies.

It's worth reminding students that mastery in differentiating requires not just memorizing formulas but understanding how they're derived and how they apply to complex functions, such as those involving compositions of functions.

Inverse Functions Calculus

When working with inverse functions in calculus, we need to keep in mind that the derivative of an inverse function can be computed using established formulas. These formulas often stem from the fact that an inverse function essentially 'reverses' the effect of the original function. For instance, the arc tangent or inverse tangent function, denoted as \( \tan^{-1} \), reverses the tangent function. This relationship lets us derive a formula for its derivative which is then applicable to any derivative involving the inverse tangent. Similarly, for the inverse cosecant or \( \csc^{-1} \), we rely on its formula for differentiation. Calculating these derivatives accurately is crucial for tasks that involve rates of change where these functions are present, such as the example exercise.

Chain Rule for Derivatives

The chain rule for derivatives is a quintessential tool in calculus, allowing us to differentiate compositions of functions. It states that if a variable \( y \) depends on \( u \) which in turn depends on \( x \) (expressed as \( y = f(u(x)) \) ), then the derivative of \( y \) with respect to \( x \) is the product of the derivative of \( y \) with respect to \( u \) and the derivative of \( u \) with respect to \( x \) (formally, \( dy/dx = (dy/du) \) \( (du/dx)\)). In our problem, we see the chain rule in action with \( u = \sqrt{x^2 - 1} \) and \( y = \tan^{-1}(u) \). Here, after finding \( d\tan^{-1}(u)/du \) and \( du/dx \), we multiply them to get \( dy/dx \) for the first term of our function. This method simplifies the process of finding derivatives of complex functions.

Trigonometric Function Derivatives

Derivatives of trigonometric functions and their inverses are a staple in any calculus toolkit. Unlike basic trigonometric functions that have straightforward derivatives, the derivatives of inverse trigonometric functions involve expressions with square roots and are derived from the relationship between the original and inverse functions. For example, the derivative of the inverse sine function \( \sin^{-1}(x) \) contains a radical in the denominator, representing the relationship to the original sine function. In the given exercise, understanding that \( \tan^{-1}\) and \( \csc^{-1}\) have specific derivative formulas - which account for their inverse nature - allows us to navigate the problem efficiently. Being comfortable with these derivatives expands the range of problems we can solve, especially when differentiating a function that involves trigonometric aspects within its composition.