Problem 35 - Thomas Calculus

Question

In Exercises \(25-36,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\sinh ^{-1}(\tan x) $$

Step-by-Step Solution

Answer

The derivative of \(y = \sinh^{-1}(\tan x)\) is \(\sec x\).

Step 1: Identify the Function and its Composition

The given function is \(y = \sinh^{-1}(\tan x)\). This is an inverse hyperbolic sine function composed with a tangent function. The derivative involves applying the chain rule.

Step 2: Derivative of Inverse Hyperbolic Sine Function

The derivative of the inverse hyperbolic sine function \(\sinh^{-1}(u)\) with respect to \(u\) is \(\frac{1}{\sqrt{u^2+1}}\). Set \(u = \tan x\).

Step 3: Derivative of the Inside Function

Find the derivative of \(u = \tan x\) with respect to \(x\). The derivative is \(\sec^2 x\).

Step 4: Apply the Chain Rule

Apply the chain rule: \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\). Substitute the derivatives from Steps 2 and 3: \(\frac{dy}{dx} = \frac{1}{\sqrt{\tan^2 x + 1}} \times \sec^2 x\).

Step 5: Simplify the Expression

Recognize that \(\tan^2 x + 1 = \sec^2 x\). So the expression simplifies to \(\frac{sec^2 x}{\sqrt{sec^2 x}} = \frac{sec^2 x}{sec x} = \sec x\). Therefore, \(\frac{dy}{dx} = \sec x\).

Key Concepts

Inverse Hyperbolic Functions

When dealing with inverse hyperbolic functions, we often explore functions that arise when you "undo" hyperbolic functions. The hyperbolic sine, \( \sinh(x)\), has an inverse function called the inverse hyperbolic sine, denoted by \( \sinh^{-1}(x)\). This is also known as "arsinh(x)".
Inverse hyperbolic functions are useful in various mathematical and engineering fields to model systems that have a hyperbola-like behavior. They offer an alternative to trigonometric functions, especially when solving differential equations.

• The inverse hyperbolic sine function domain is all real numbers.
• Its range is also all real numbers, similar to basic algebraic functions.
• The derivative for \( \sinh^{-1}(x)\) with respect to \( x \) simplifies the evaluation of integrals too.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. When you have a function that is composed of two functions, such as \( y = \sinh^{-1}(\tan x) \), the chain rule allows you to differentiate it step by step.
Think of it as a way to "peel away" the layers of a function until you're able to work with basic derivatives. The chain rule formula states: \[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]where \( y = f(u) \) and \( u = g(x) \).

• First, identify the inner function (\( u = g(x) \), which in this case is \( u = \tan(x) \)).
• Next, find the derivative of the outer function with respect to the inner function ( \( y = \sinh^{-1}(u) \)).
• Finally, multiply these derivatives to find \( \frac{dy}{dx} \).

Trigonometric Functions

Trigonometric functions are essential to calculus and many areas of mathematics. They describe relationships in right-angled triangles, among other uses. The function \( \tan(x) \) is one such function and is defined as the ratio of the sine and cosine functions: \[\tan(x) = \frac{\sin(x)}{\cos(x)}\]A hallmark of calculus is understanding how to work with these trigonometric functions, especially when finding derivatives or integrals.

• Common derivatives include \( \frac{d}{dx}\tan(x) = \sec^2(x) \).
• The identity \( \tan^2(x) + 1 = \sec^2(x)\) helps simplify expressions, just like in our exercise here.

Mastery of these identities and derivatives is crucial for problem-solving in calculus, enabling one to convert complex expressions into simpler forms.

Calculus Problem Solving

Calculus problem-solving often entails a systematic approach to breaks down complicated problems into manageable sections. Here, solving the derivative of \( y = \sinh^{-1}(\tan x) \) requires applying specific core calculus concepts neatly:

• Recognize and decompose the function into layers, like the inverse hyperbolic and trigonometric functions.
• Use appropriate formulas, such as the chain rule, to find derivatives efficiently.
• Look for algebraic simplifications, like the identity \( \tan^2(x) + 1 = \sec^2(x)\), to streamline expressions.

As you practice more, you'll better anticipate the steps needed and quickly identify which techniques to apply. Keeping a keen eye for patterns, identities, and simplifications in each problem can save time and improve accuracy in calculus exercises.