Question

In Exercises 83 and \(84,\) use the equation of the tractrix \(y=a \operatorname{sech}^{-1} \frac{x}{a}-\sqrt{a^{2}-x^{2}}, \quad a>0\) Find \(d y / d x\).

Short answer

The derivative \(dy/dx\) of the function \(y = a \operatorname{sech}^{-1} \frac{x}{a} - \sqrt{a^2 - x^2}\) is: \( -\frac{x}{a^2 \sqrt{1-(x^2/a^2)}} - \frac{x}{\sqrt{a^2 - x^2}} \).

Step-by-step solution

Identify the function's parts

Separate the function into two simpler functions, \(f(x) = a \operatorname{sech}^{-1} \frac{x}{a}\) and \(g(x) = -\sqrt{a^2 - x^2}\). Then apply the sum/difference rule to find the derivative.

Derive \(f(x)\)

To find the derivative of \(f(x)\), apply the chain rule. The chain rule is a formula to compute the derivative of a composite function: \((f(g(x)))' = f'(g(x)) ∗ g'(x)\). The derivative of \(\operatorname{sech}^{-1} y\) is \(-1/(|y|\sqrt{1-y^2})\), where \(y = \frac{x}{a}\), and the derivative of \(y = \frac{x}{a}\) is \(1/a\). Subtract it from 1.

Derive \(g(x)\)

To find the derivative of \(g(x)\), apply the chain rule. The derivative of \(-\sqrt{1-y^2}\) is \(y/\sqrt{1-y^2}\), where \(y = a^2 - x^2\), and the derivative of \(y = a^2 - x^2\) is \(-2x\). Combine these to get the derivative of \(g(x)\).

Combine the derivatives

Combine the derivatives from steps 2 and 3 to get the derivative of the original function. Apply the rule \(f(g(x))' = f'(g(x)) \cdot g'(x)\).

Key concepts

Chain Rule

When studying calculus, one of the fundamental tools you encounter is the chain rule. It's a formula used to compute the derivative of a composite function. Suppose you have functions f and g, where f is a function of g(x), i.e., f(g(x)). Instead of taking the derivatives separately, the chain rule allows us to differentiate the composite function smoothly.

The formula is expressed as \[ (f(g(x)))' = f'(g(x)) \cdot g'(x) \. \] This means that to find the derivative of the composite function, you take the derivative of the outer function, evaluated at the inner function (f'(g(x))), and multiply it by the derivative of the inner function (g'(x)).

In our exercise, the chain rule is applied to both terms separately, with each term considered as a composite function. For example, in step 2, we differentiate the inverse hyperbolic secant function - which is itself a function of (x/a) - by applying the chain rule.

Implicit Differentiation

While the standard procedure of differentiation works well with explicit functions where y is given directly in terms of x, implicit differentiation is useful for equations where y is not isolated. It allows us to find the derivative dy/dx without having to solve for y explicitly.

This doesn't come to the forefront in our current problem since the tractrix equation is explicitly given in terms of y. However, understanding implicit differentiation expands one's ability to tackle derivatives of functions that might not be easily solvable for y.

The general approach involves differentiating both sides of an equation with respect to x and then solving for dy/dx. In doing so, we apply the chain rule whenever we differentiate a term involving y, since y is a function of x. Although not used in the explicit differentiation of our tractrix function, implicit differentiation underpins understanding derivatives where y and x are deeply intertwined.

Hyperbolic Functions

In the world of calculus, hyperbolic functions come into play as analogs of the trigonometric functions, but for hyperbolic geometry. These include hyperbolic sine (\text{sinh}), hyperbolic cosine (\text{cosh}), and their inverses, among others.

The hyperbolic functions are related to the exponential function, and they have many useful properties that mirror the familiar trigonometric functions. In our exercise, we deal with the inverse hyperbolic secant, \( \operatorname{sech}^{-1} \), which is the inverse function of the hyperbolic secant (\text{sech}). The derivative of an inverse hyperbolic function, like other inverse functions, requires knowledge of the function it is inverting.

The derivative of \( \operatorname{sech}^{-1} y \) is given by \[ -\frac{1}{|y|\sqrt{1-y^2}} \. \] This formula is a critical piece in finding the derivative of the tractrix equation given in the exercise. Understanding how to differentiate hyperbolic functions, especially their inverse forms, is essential for solving calculus problems involving these unique functions.