Problem 287 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for the given function. $$y=\cot ^{-1} \sqrt{4-x^{2}}$$

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = \frac{x}{(5-x^2)\sqrt{4-x^2}} \)

Step 1: Identify the Function and Rewrite

The function given is \( y = \cot^{-1}(\sqrt{4-x^2}) \). First, note that \( \cot^{-1} \) is the inverse cotangent function. We'll need to apply the chain rule and trigonometric derivative rules to find \( \frac{d y}{d x} \).

Step 2: Differentiate the Outside Function

The outer function is \( \cot^{-1}(u) \) where \( u = \sqrt{4-x^2} \). The derivative of \( \cot^{-1}(u) \) is \( -\frac{1}{1+u^2} \). Thus, \( \frac{d}{du} \cot^{-1}(u) = -\frac{1}{1+u^2} \).

Step 3: Differentiate the Inside Function

The inside function is \( u = \sqrt{4-x^2} \). To differentiate, write as \( u = (4-x^2)^{1/2} \). The derivative is \( \frac{du}{dx} = \frac{1}{2}(4-x^2)^{-1/2}(-2x) \), which simplifies to \( -\frac{x}{\sqrt{4-x^2}} \).

Step 4: Apply the Chain Rule

Now apply the chain rule: \( \frac{d y}{d x} = \frac{d y}{du} \cdot \frac{du}{dx} \). Substitute the derivatives obtained, \( \frac{d y}{dx} = -\frac{1}{1+u^2} \cdot -\frac{x}{\sqrt{4-x^2}} \).

Step 5: Substitute Back for \(u\)

Recall that \( u = \sqrt{4-x^2} \). Substitute \( u \) back into the derivative expression: \( \frac{d y}{d x} = \frac{x}{(1+(4-x^2))\sqrt{4-x^2}} \).

Step 6: Simplify the Expression

Simplify the expression: \( 1 + (4-x^2) = 5-x^2 \). Thus, \( \frac{d y}{d x} = \frac{x}{(5-x^2)\sqrt{4-x^2}} \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus used for differentiating compositions of functions. It may seem daunting at first, but breaking it down makes it manageable.
For a function expressed as a composite, say \( f(g(x)) \), the derivative is taken by differentiating the outer function \( f \) with respect to its inner function \( g(x) \), and then multiplying by the derivative of the inside function \( g \) with respect to \( x \). The formal definition is \( \frac{d}{dx} f(g(x)) = f'(g(x)) \, g'(x) \).

• Identify the outer and inner functions. Here, the outer function was \( \cot^{-1}(u) \) and the inner \( \sqrt{4-x^2} \).
• Differentiating the outer function with respect to the inner simply means replacing its 'argument' with the inner function. In this case, \( \frac{d}{du} \cot^{-1} (u) \).
• Next, multiply this by the derivative of the inner function. These steps combine to follow the chain rule.

This simple multiplication provides the powerful capability to deal with complicated expressions by breaking them down into easily manageable parts.

Trigonometric Derivatives

Trigonometric derivatives give us the rate of change of trigonometric functions. These are connecting blocks in calculus, especially the less common inverse trigonometric functions which could seem difficult.
Derivatives of trigonometric functions like \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \) are more straightforward through standard identities. However, derivatives of inverse trigonometric functions like \( \cot^{-1}(x) \) are also well-defined. The derivative of \( \cot^{-1}(u) \) is \( -\frac{1}{1+u^2} \).

• Recognize the inverse trigonometric function: In our case, \( \cot^{-1} \) plays a role as the outer layer of our composite.
• Use known standard derivatives: These help simplify the differentiation steps, turning more complex expressions into direct applications of known rules.

Leveraging these known derivatives effectively can accelerate finding solutions and deepening understanding of calculus principles.

Inverse Trigonometric Functions

Inverse trigonometric functions are the reverse processes of standard trigonometric functions. They take a known ratio and find the corresponding angle that stems from these standard trigonometric functions.
The inverse function \( \cot^{-1}(x) \) provides the angle whose cotangent is \( x \). Dealing with these functions often requires precise manipulation and a strong grasp of how they "undo" the normal trigonometric operations.

• Understand the function's domain and range: This helps when differentiating, as knowing where the function applies can guide simplifying expressions.
• Consider any trigonometric identities: Relationships like \( \cot(x) = \frac{1}{\tan(x)} \) can assist when differentiating more complicated expressions.

Knowing these functions inside out makes tackling exercises involving them much easier and engaging, providing smoother paths through the challenges of calculus.