Problem 31 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative is \(-\operatorname{sech}^{-1} x\).

Step 1: Identify the Components

We need to differentiate the function \( y = \cos^{-1} x - x \cdot \operatorname{sech}^{-1} x \). This function consists of two parts: \( \cos^{-1} x \), the inverse cosine function, and \( x \cdot \operatorname{sech}^{-1} x \) which is a product of \( x \) and the inverse hyperbolic secant function.

Step 2: Differentiate \( \cos^{-1} x \)

The derivative of \( \cos^{-1} x \) with respect to \( x \) is \( -\frac{1}{\sqrt{1-x^2}} \).

Step 3: Use Product Rule for \( x \cdot \operatorname{sech}^{-1} x \)

To differentiate \( x \cdot \operatorname{sech}^{-1} x \), apply the product rule: if \( u = x \) and \( v = \operatorname{sech}^{-1} x \), then the derivative is \( u'v + uv' \).

Step 4: Differentiate \( u = x \)

The derivative of \( u = x \) with respect to \( x \) is 1.

Step 5: Differentiate \( v = \operatorname{sech}^{-1} x \)

The derivative of \( \operatorname{sech}^{-1} x \) with respect to \( x \) is \( -\frac{1}{x\sqrt{1-x^2}} \).

Step 6: Combine Product Rule Components

Using the product rule formula, \( u'v + uv' = 1 \cdot \operatorname{sech}^{-1} x + x \cdot \left(-\frac{1}{x\sqrt{1-x^2}}\right) = \operatorname{sech}^{-1} x - \frac{1}{\sqrt{1-x^2}} \).

Step 7: Derivative of the Entire Function

Thus, the derivative \( y' \) is: \[-\frac{1}{\sqrt{1-x^2}} - (\operatorname{sech}^{-1} x - \frac{1}{\sqrt{1-x^2}}) = -\operatorname{sech}^{-1} x \].

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions represent angles given the value of the trigonometric ratio. In our exercise, we deal with the inverse cosine function, denoted as \( \cos^{-1}(x) \). This function provides the angle whose cosine is \( x \). It is important in problems involving angles and lengths in right triangles.

When differentiating inverse trigonometric functions, special derivatives are used. For \( \cos^{-1}(x) \), the derivative with respect to \( x \) is \(-\frac{1}{\sqrt{1-x^2}}\). Here’s what you need to remember:

• The derivative is negative, indicating the inverse cosine function is decreasing.
• The expression \( \sqrt{1-x^2} \) comes from geometric relationships inherent in trigonometry.

Using these derivatives correctly is crucial in calculus when finding slopes or rates of change related to angles.

Product Rule

The product rule is a fundamental concept in calculus used to differentiate products of two functions. If you have a function, say \(y = uv\), where both \(u\) and \(v\) are functions of \(x\), the derivative of \(y\) is given by \(u'v + uv'\).

This rule is applied in our exercise to differentiate the product \(x \cdot \operatorname{sech}^{-1}(x)\). Here's how it works:

• Differentiating \(u = x\) gives \(u' = 1\).
• Differentiating \(v = \operatorname{sech}^{-1}(x)\) gives \(v' = -\frac{1}{x\sqrt{1-x^2}}\).
• Inserting into the formula, we compute \(u'v + uv' = 1 \cdot \operatorname{sech}^{-1}(x) + x \cdot \left(-\frac{1}{x\sqrt{1-x^2}}\right)\).

It’s vital to be comfortable with the product rule to tackle scenarios where two or more functions are multiplied, helping you compute derivatives efficiently.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are analogs of inverse trigonometric functions, but relate to hyperbolas instead of circles. In this exercise, the function \( \operatorname{sech}^{-1}(x) \) is the inverse hyperbolic secant.

These functions provide angles or arguments whose hyperbolic ratios are specific values. Differentiating inverse hyperbolic functions often involves unique formulas distinct from their trigonometric counterparts.

• The derivative of \(\operatorname{sech}^{-1}(x)\) is \(-\frac{1}{x\sqrt{1-x^2}}\).
• This derivative reflects how the inverse hyperbolic secant changes with changes in \(x\).

Understanding the derivatives of inverse hyperbolic functions is crucial for tasks involving growth models, physics, or calculus problems related to hyperbolic functions. Being familiar with these derivatives allows you to solve complex differentiation problems with confidence.