Problem 33 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \operatorname{arccot}(\sqrt{x}) $$

Step-by-Step Solution

Answer

The derivative is \(-\frac{1}{2\sqrt{x}(1 + x)}\).

Step 1: Recall the Derivative of Arccotangent

The derivative of the arccotangent function with respect to a variable is given by \( \frac{d}{dx} [\operatorname{arccot}(u)] = -\frac{1}{1 + u^2} \cdot \frac{du}{dx} \). We'll apply this to our function \( \operatorname{arccot}(\sqrt{x}) \).

Step 2: Identify Inner Function and its Derivative

We identify \( u = \sqrt{x} \). Therefore, the derivative of \( u \) with respect to \( x \) is \( \frac{du}{dx} = \frac{d}{dx}[x^{1/2}] = \frac{1}{2\sqrt{x}} \).

Step 3: Apply the Chain Rule

Using the chain rule, substitute \( u = \sqrt{x} \) and \( \frac{du}{dx} \) into the derivative of arccotangent formula: \[ \frac{d}{dx}[\operatorname{arccot}(\sqrt{x})] = -\frac{1}{1 + (\sqrt{x})^2} \times \frac{1}{2\sqrt{x}} = -\frac{1}{1 + x} \times \frac{1}{2\sqrt{x}} \].

Step 4: Simplify the Expression

Combine the fractions: \[ \frac{d}{dx}[\operatorname{arccot}(\sqrt{x})] = -\frac{1}{2\sqrt{x}(1 + x)} \]. This is the simplified form of the derivative.

Key Concepts

Arccotangent Function

The arccotangent function, commonly written as \( \operatorname{arccot}(x) \), is the inverse of the cotangent function. In the context of trigonometry, the cotangent of an angle in a right triangle is the ratio of the adjacent side to the opposite side. When we take the arccotangent, we are essentially asking: "What angle does this ratio correspond to?"

Some important properties of the arccotangent function include its domain and range:

• Domain: all real numbers \( (-\infty, \infty) \)
• Range: \((0, \pi)\)

This means that the output of the arccotangent is always between 0 and \( \pi \), unlike the tangent or cotangent functions, which can take on any real value.

Understanding the derivative of the arccotangent function is crucial; it is expressed as \( \frac{d}{dx}[\operatorname{arccot}(u)] = -\frac{1}{1 + u^2} \). This formula shows that the derivative is always negative, as the numerator \(-1\) is constant while the denominator \((1 + u^2)\) is always positive for any real \(u\).

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. In simple terms, if you have a function within another function, the chain rule helps you differentiate them step by step.

To use the chain rule, first identify the outer and inner functions. In our example, the outer function is the arccotangent, and the inner function is \( \sqrt{x} \).

The chain rule interlinks these functions as follows: first differentiate the outer function \( \operatorname{arccot}(u) \) with respect to the inner function \( u \), and then multiply by the derivative of the inner function \( u = \sqrt{x} \). Hence, the chain rule states:

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

Incorporating this into our problem, we first find the derivative of the arccotangent function with respect to its argument and then multiply it by the derivative of \( \sqrt{x} \). This step-by-step process simplifies handling complex composite functions.

Simplified Derivative Expression

Simplifying a derivative expression involves reducing it to its most compact and understandable form. Once the chain rule has been applied, we are often left with a complex expression. Our goal is to simplify it to make further mathematical calculations easier.

In our example, after applying the chain rule, we arrived at \(-\frac{1}{1 + x} \times \frac{1}{2\sqrt{x}}\). To simplify, we combine these separate parts into a single rational expression. This can involve multiplying terms together, as fractions or radicals need to be handled:

The simplified derivative turns into \(-\frac{1}{2\sqrt{x}(1 + x)}\). This expression is easier to work with in mathematical calculations or graphing because:

• It's a single fraction
• No extra steps or separate factors are involved
• The radicals are properly managed in the denominator

Such simplification ensures that anyone working with the derivative of complex functions can follow through calculations smoothly and without unnecessary complications.