Problem 68 - Thomas Calculus

Question

In Exercises \(49-70\) , find the derivative of \(y\) with respect to the appropriate variable. $$ y=\cot ^{-1} \frac{1}{x}-\tan ^{-1} x $$

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} \) is \( \frac{x^2 - 1}{x^2 + 1} \).

Step 1: Recall the Derivative Formulas

The derivative of the inverse cotangent function, \( \cot^{-1}(u) \), is \( \frac{-1}{1+u^2} \frac{du}{dx} \). Similarly, the derivative of the inverse tangent function, \( \tan^{-1}(u) \), is \( \frac{1}{1+u^2} \frac{du}{dx} \). We will apply these formulas to find \( \frac{dy}{dx} \).

Step 2: Find the Derivative of \( \cot^{-1}\left(\frac{1}{x}\right) \)

Let \( u = \frac{1}{x} \). Then \( \frac{du}{dx} = -\frac{1}{x^2} \). Using the derivative formula for \( \cot^{-1}(u) \), we have:\[\frac{d}{dx} \left( \cot^{-1}\left(\frac{1}{x}\right) \right) = \frac{-1}{1+\left(\frac{1}{x}\right)^2} \left(-\frac{1}{x^2}\right) = \frac{1}{1 + \frac{1}{x^2}} \cdot \frac{1}{x^2}\]Simplifying, we get:\[\frac{x^2}{x^2 + 1}\]

Step 3: Find the Derivative of \( \tan^{-1}(x) \)

Use the derivative formula for the inverse tangent function:\[\frac{d}{dx} (\tan^{-1}(x)) = \frac{1}{1+x^2}\]

Step 4: Combine the Results

The derivative of the given function \( y = \cot^{-1}\left(\frac{1}{x}\right) - \tan^{-1}(x) \) is the difference of the two derivatives computed:\[\frac{dy}{dx} = \frac{x^2}{x^2 + 1} - \frac{1}{1+x^2}\]Simplifying:\[\frac{x^2}{x^2 + 1} - \frac{1}{x^2 + 1} = \frac{x^2 - 1}{x^2 + 1}\]

Step 5: Conclusion

Therefore, the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{x^2 - 1}{x^2 + 1} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the standard trigonometric functions. They help us solve equations where the angle is the unknown we need to find. The inverse cotangent, denoted as \( \cot^{-1}(x) \), and the inverse tangent, denoted as \( \tan^{-1}(x) \), are examples of these functions.
Understanding their properties and derivatives is essential for solving calculus problems involving inverse trigonometric functions.

• \( \cot^{-1}(x) \) gives us the angle whose cotangent is \( x \).
• \( \tan^{-1}(x) \) gives us the angle whose tangent is \( x \).

These functions have specific derivative formulas which play a crucial role when differentiating expressions involving them.

Derivative Formulas

Derivative formulas are like handy rules that help us differentiate functions easily. When it comes to inverse trigonometric functions, remembering their derivative formulas is vital.For the inverse cotangent function \( \cot^{-1}(u) \), the derivative is:\[\frac{d}{dx} \left( \cot^{-1}(u) \right) = \frac{-1}{1+u^2} \cdot \frac{du}{dx}\]Similarly, the derivative for the inverse tangent function \( \tan^{-1}(u) \) is:\[\frac{d}{dx} \left( \tan^{-1}(u) \right) = \frac{1}{1+u^2} \cdot \frac{du}{dx}\]

• These formulas are helpful as they allow substitution and further simplification when taking derivatives.
• In the original problem, identifying \( u \) and computing \( \frac{du}{dx} \) were crucial steps for applying these formulas.

Learning these formulas helps save time when tackling calculus problems.

Calculus Problem Solving

Solving calculus problems often involves applying derivatives and simplifications to reach the final answer efficiently. Here are a few key steps from the original problem solution:1. **Understand the Problem:** We need to find the derivative of a function involving inverse trigonometric functions.2. **Apply Derivative Formulas:** Use the derivative formulas for \( \cot^{-1}\left(\frac{1}{x}\right) \) and \( \tan^{-1}(x) \). We find this by setting substitution for variables and differentiating accordingly.3. **Simplify the Result:** After applying the derivatives, the expression might look complex. Simplifying it gives a more straightforward and more comprehensible result.For the original function, the derivative got simplified from:\[\frac{dy}{dx} = \frac{x^2}{x^2 + 1} - \frac{1}{1+x^2}\]to:\[\frac{x^2 - 1}{x^2 + 1}\]

• This step is crucial as it not only finds the derivative but optimizes its form, which might be necessary for further computations or analyses.

By following these problem-solving steps effectively, understanding and solving calculus problems becomes much easier. Remember, practice is key to mastering these methods!