Problem 9 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Compute \(\frac{d}{d x} \frac{\sin ^{-1} x}{\cos ^{-1} x} .\) Is it the same as \(\frac{d}{d x} \tan ^{-1} x ?\)

Step-by-Step Solution

Answer

No, the expression \(\frac{d}{d x} \frac{\sin ^{-1} x}{\cos ^{-1} x}\) is not the same as \(\frac{d}{d x}\tan^{-1}x\). Their derivatives deliver different values.

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

To differentiate \(\frac{\sin ^{-1} x}{\cos ^{-1} x}\), apply the quotient rule for differentiation which is \(\frac{d}{dx}\frac{u}{v} = \frac{vu' - uv'}{v^2}\). Let \(\sin ^{-1} x\) be \(u\) and \(\cos ^{-1} x\) be \(v\). Next is to find \(u'\) and \(v'\), the derivatives of \(u\) and \(v\) respectively. The derivative of \(\sin ^{-1} x\) is \(\frac{1}{\sqrt{1 - x^2}}\) and the derivative of \(\cos ^{-1} x\) is \(-\frac{1}{\sqrt{1 - x^2}}\). Now replace \(u\), \(v\), \(u'\) and \(v'\) in the quotient rule equation.

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

Next, differentiate \(\tan ^{-1} x\). The derivative of \(\tan ^{-1} x\) is \(\frac{1}{1 + x^2}\).

Step 3: Compare the results

Compare the result from step 1 and step 2 to determine if they are the same. This will involve simplifying the derivative obtained in step 1 before comparing.

Key Concepts

Quotient Rule

The quotient rule is a critical technique in calculus used for finding the derivative of a ratio of two functions. When you have a function that looks like a fraction where both the numerator and denominator are functions of the same variable, the quotient rule is your tool. In this instance, the function is \( \frac{\sin^{-1} x}{\cos^{-1} x} \). The quotient rule formula is:

• \( \frac{d}{dx}\frac{u}{v} = \frac{vu' - uv'}{v^2} \)

Here, \(u\) is \(\sin^{-1}x\) and \(v\) is \(\cos^{-1}x\). Both \(u\) and \(v\) are inverse trigonometric functions (more on that below), making this particular derivative a little more complex than a basic polynomial function.
To apply the quotient rule, determine the derivatives \(u'\) and \(v'\) first. Then insert them back into the formula. This allows you to compute the derivative of the function as a whole.

Inverse Trigonometric Functions

Inverse trigonometric functions are the reverse operations of the regular trigonometric functions. They are solutions for angles when the values of the trigonometric functions (like sine, cosine, etc.) are known. Functions like \(\sin^{-1}x\) and \(\cos^{-1}x\) are called "arc functions" since they determine the arc, or angle, whose sine or cosine is \(x\).
The main inverse functions include:

• \(\sin^{-1}x\), the arcsine
• \(\cos^{-1}x\), the arccosine
• \(\tan^{-1}x\), the arctangent

Each of these has a specific derivative which you'll need often in calculus problems involving derivatives. In the original exercise, the derivative of \(\sin^{-1}x\) is \(\frac{1}{\sqrt{1 - x^2}}\) and for \(\cos^{-1}x\), it's \(-\frac{1}{\sqrt{1 - x^2}}\). Recognize and remember these derivatives as they frequently appear in differentiation tasks.

Differentiation

Differentiation is a fundamental operation in calculus. It involves finding the derivative, which tells us how a function behaves when there's an infinitesimally small change in the variable. Differentiating a function is like finding the slope of its graph at any point.
The process can get tricky when dealing with complex or composite functions, such as products, quotients, or compositions of inverse trigonometric functions. This is why rules like the quotient rule or the ability to differentiate inverse trigonometric functions are essential.

• Simple functions, like polynomials, have straightforward derivatives.
• For more complex functions like \(\frac{\sin^{-1} x}{\cos^{-1} x}\), rules such as the quotient rule are applied.

By practicing these differentiation techniques, you build a foundation for tackling more intricate problems in calculus.