Problem 36 - University Calculus: Early Transcendentals

Question

Find \(d y\). $$y=\cot ^{-1}\left(\frac{1}{x^{2}}\right)+\cos ^{-1} 2 x$$

Step-by-Step Solution

Answer

\(dy = \left(\frac{2x}{x^4+1} - \frac{2}{\sqrt{1-4x^2}}\right)dx\)

Step 1: Differentiate cot^(-1)(1/x^2)

Let \(u = 1/x^2 = x^{-2}\). Then \(\frac{du}{dx} = -2x^{-3}\).
\(\frac{d}{dx}[\cot^{-1}(u)] = \frac{-1}{1 + u^2} \cdot \frac{du}{dx} = \frac{-1}{1 + 1/x^4} \cdot (-2/x^3) = \frac{2/x^3}{1 + 1/x^4} = \frac{2x}{x^4 + 1}\)

Step 2: Differentiate cos^(-1)(2x)

\(\frac{d}{dx}[\cos^{-1}(2x)] = \frac{-1}{\sqrt{1 - 4x^2}} \cdot 2 = \frac{-2}{\sqrt{1 - 4x^2}}\)

Step 3: Write dy

\(dy = \left(\frac{2x}{x^4 + 1} - \frac{2}{\sqrt{1 - 4x^2}}\right) dx\)

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, like the inverse cotangent and inverse cosine, are fundamental in calculus. These functions allow us to work backward from the trigonometric ratio to find angles, which is crucial in many applications, such as physics and engineering. When you see notations such as \( \cot^{-1}(x) \), it refers to the angle whose cotangent is \( x \). This is different from their direct trigonometric counterparts, which directly calculate the ratios from an angle.
When differentiating inverse trigonometric functions, specific formulas are used. These formulas are derived from the fact that these functions, although different from their trigonometric counterparts, still have distinct relationships and rates of change.

Chain Rule

The Chain Rule is a fundamental tool in calculus used to differentiate composite functions. When a function is nested inside another, such as \( y = \cot^{-1}(\frac{1}{x^2}) \), where the inner function \( u = \frac{1}{x^2} \), the Chain Rule helps break it down.
By applying the Chain Rule, we take the derivative of the outer function evaluated at the inner function and multiply it by the derivative of the inner function. For example, knowing \( \frac{d}{dx} \cot^{-1}(u) = -\frac{1}{1 + u^2} \), and having already determined \( \frac{du}{dx} = -\frac{2}{x^3} \), the product yields the derivative of the composite function. This method allows us to simplify otherwise complex differentiation tasks.

Derivative Formula

The derivative formula for inverse trigonometric functions is essential for correctly calculating derivatives. For instance, the derivative of the inverse cotangent function is defined as \( \frac{d}{dx} \cot^{-1}(u) = -\frac{1}{1 + u^2} \cdot \frac{du}{dx} \). This formula illustrates how the rate of change is influenced by both the trigonometric function itself and the change in the inner function.
These formulas are usually memorized and then applied directly, but understanding the underlying principles can immensely enhance comprehension. They result from the fundamental chain of calculations that link tangent, cotangent, and their inverses. The use of a derivative formula simplifies the calculation, making it a standard procedure in calculus problem-solving.

Mathematical Problem-Solving

Mathematical problem-solving involves applying concepts and strategies to arrive at a solution. In calculus, this often means using differentiation rules like the Chain Rule and recognizing when to apply derivative formulas.
In our original exercise, solving the problem required multiple steps: first setting up the equation, recognizing inverse trigonometric functions, differentiating using appropriate formulas, and finally applying the Chain Rule for composite functions.
Problem-solving in mathematics often follows a systematic approach: identifying the functions involved, determining applicable differentiation techniques, calculating derivatives step-by-step, and finally simplifying the results for clarity.