Problem 52 - University Calculus: Early Transcendentals

Question

Use the identity $$\cot ^{-1} u=\frac{\pi}{2}-\tan ^{-1} u$$ to derive the formula for the derivative of \(\cot ^{-1} u\) in Table 3.1 from the formula for the derivative of \(\tan ^{-1} u\)

Step-by-Step Solution

Answer

The derivative of \(\cot^{-1} u\) is \(-\frac{1}{1+u^2}\).

Step 1: Recall the Given Identity

We are given the identity \(\cot^{-1} u = \frac{\pi}{2} - \tan^{-1} u\). This identity expresses the inverse cotangent function in terms of inverse tangent.

Step 2: Derivative of the Identity Equation

Differentiate both sides of the equation \(\cot^{-1} u = \frac{\pi}{2} - \tan^{-1} u\) with respect to \(u\). The derivative of a constant, \(\frac{\pi}{2}\), is 0, so we get: \(\frac{d}{du}\left(\cot^{-1} u\right) = -\frac{d}{du}\left(\tan^{-1} u\right)\).

Step 3: Derivative of \(\tan^{-1} u\)

Recall the derivative formula for \(\tan^{-1} u\), which is \(\frac{d}{du}\left(\tan^{-1} u\right) = \frac{1}{1+u^2}\).

Step 4: Substitute the Derivative

Substitute the derivative of \(\tan^{-1} u\) from Step 3 into the derivative equation from Step 2: \(-\frac{d}{du}\left(\tan^{-1} u\right) = -\frac{1}{1+u^2}\).

Step 5: Conclusion

Thus, the formula for the derivative of \(\cot^{-1} u\) is \(\frac{d}{du}\left(\cot^{-1} u\right) = -\frac{1}{1+u^2}\).

Key Concepts

Derivative Rules

Derivative rules are fundamental tools in calculus that allow us to find the rate of change or slope of a curve at any given point. These rules provide the formulas we need to efficiently calculate derivatives without having to use the limit definition repeatedly. Some of the most common derivative rules include:

• The Product Rule: Used when differentiating products of two functions, given by \( (uv)' = u'v + uv' \).
• The Quotient Rule: Used for functions that are divided by one another, described by \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
• The Chain Rule: Essential for differentiating composite functions, stated as \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).

Derivatives are a core concept of calculus and are widely used in fields ranging from physics to economics. Understanding the standard rules is crucial for more advanced applications.

Inverse Cotangent

The inverse cotangent function, written as \( \cot^{-1}(u) \), is the arc cotangent of a number. It represents the angle whose cotangent is that number. Similar to other inverse trigonometric functions, it's the reverse operation of taking the cotangent of an angle.

To understand the inverse cotangent more deeply, it's helpful to visualize or rethink its range. Unlike cotangent, which is defined for many outputs, inverse cotangent is typically restricted to output angles between 0 and \( \pi \). This restriction helps in making the function single-valued and simplifies its use in calculus.

A key property of inverse cotangent is its relationship with inverse tangent, expressed as \( \cot^{-1} u = \frac{\pi}{2} - \tan^{-1} u \). This identity is crucial when finding derivatives, as we've seen in transforming the derivative of the inverse tangent to that of the inverse cotangent. The derivative formula for \( \cot^{-1}(u) \) is given by \( \frac{d}{du}(\cot^{-1}u) = -\frac{1}{1+u^2} \). This formula is particularly useful in calculus for finding rates of change involving arc cotangent.

Inverse Tangent

The inverse tangent function, denoted as \( \tan^{-1}(u) \) or arctan, is one of the most commonly encountered inverse trigonometric functions. It tells us the angle whose tangent is the given number, effectively reversing the tangent operation.

• Range: The function confers angles from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). This restriction ensures the function is single-valued.
• Derivative: The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1+u^2} \). This simple expression is quite powerful and finds applications in various calculus problems.
• Identity: The identity \( \cot^{-1}(u) = \frac{\pi}{2} - \tan^{-1}(u) \) illuminates the relationship between inverse tangent and inverse cotangent.

With these attributes, the inverse tangent helps solve trigonometric equations and compute integrals and derivatives in calculus effectively. Understanding its derivative is especially useful, as it forms a basis for many other calculus concepts.