Problem 23 - Essential Calculus Early Transcendentals

Question

Find the derivative of the function. Simplify where possible. $$h(t)=\cot ^{-1}(t)+\cot ^{-1}(1 / t)$$

Step-by-Step Solution

Answer

The derivative of the function is 0.

Step 1: Recall the derivative of the inverse cotangent function

The derivative of the inverse cotangent function, \(\cot^{-1}(x)\), with respect to \(x\) is given by \(\frac{d}{dx} \cot^{-1}(x) = -\frac{1}{1+x^2}\). We will use this formula to differentiate each term of the given function.

Step 2: Differentiate the first term \(\cot^{-1}(t)\)

The derivative of the first term \(\cot^{-1}(t)\) is obtained using \(\frac{d}{dt} \cot^{-1}(t) = -\frac{1}{1+t^2}\).

Step 3: Differentiate the second term \(\cot^{-1}(1/t)\) using Chain Rule

Apply the chain rule to differentiate \(\cot^{-1}(1/t)\). Let \( u = 1/t \), then the derivative of \(\cot^{-1}(u)\) with respect to \(u\) is \(-\frac{1}{1+u^2}\). The derivative of \(u\) with respect to \(t\) is \(-1/t^2\). So, \(\frac{d}{dt} \cot^{-1}(1/t) = \frac{1/t^2}{1+(1/t)^2} = \frac{1}{t^2 + 1}\).

Step 4: Combine the derivatives from steps 2 and 3

Combining the results from step 2 and step 3, the derivative of \(h(t) = \cot^{-1}(t) + \cot^{-1}(1/t)\) becomes: \(-\frac{1}{1+t^2} + \frac{1}{t^2+1}\).

Step 5: Simplify the expression

Notice that the two derivatives from step 4 cancel each other since the expressions \(-\frac{1}{1+t^2}\) and \(\frac{1}{t^2+1}\) are identical in magnitude but have opposite signs. Therefore, the sum results in zero. Thus, the derivative of the function is 0.

Key Concepts

Inverse Cotangent

When thinking about trigonometric functions, you're probably familiar with sine, cosine, and tangent. But have you ever heard of the inverse cotangent? The inverse cotangent function, denoted as \( \cot^{-1}(x) \), is a function that reverses the cotangent's operation. In simpler terms, if you know the cotangent of an angle, using the inverse cotangent helps you find that angle.

Inverse trigonometric functions like \( \cot^{-1}(x) \) are widely used in calculus to simplify and solve equations involving angles. One thing to remember about \( \cot^{-1}(x) \) is its range. It specifically gives you angles from \( 0 \) to \( \pi \), excluding \( \frac{\pi}{2} \) where the cotangent function is undefined. This choice ensures that the output is unique and provides a clear representation of the angle.

Chain Rule

Differentiation is all about finding rates of change, and sometimes functions have other functions inside them. This is where the chain rule comes in handy. The chain rule is crucial for differentiating composite functions—functions within functions.

In simple terms, if you have a function \( f(g(x)) \), the chain rule says you differentiate \( f \) with respect to \( g \), then multiply by the derivative of \( g \) with respect to \( x \). Consider differentiating \( \cot^{-1}(1/t) \), a composite function where the external function is \( \cot^{-1} \) and the internal function is \( 1/t \):

• First, differentiate \( \cot^{-1}(u) \) where \( u = 1/t \). This gives \( -\frac{1}{1+u^2} \).
• Next, differentiate \( 1/t \) to get \( -1/t^2 \).
• Finally, multiply them: \( \left(-\frac{1}{1+(1/t)^2}\right) \left(-\frac{1}{t^2}\right) = \frac{1}{t^2+1} \).

Thus, the chain rule helps find derivatives when confronted with nested functions.

Differentiation

Differentiation is one of the key operations in calculus, letting us find how functions change and behave. When you differentiate a function, you are essentially asking, "What is the slope or rate of change at any point?"

In the exercise given, you are tasked with finding the derivative of a function involving inverse cotangents. Both components of the function need to be differentiated separately and then combined. For \( h(t) = \cot^{-1}(t) + \cot^{-1}(1/t) \), here's the differentiation process:

• The derivative of \( \cot^{-1}(t) \) is \( -\frac{1}{1+t^2} \).
• Using the chain rule, the derivative for \( \cot^{-1}(1/t) \) becomes \( \frac{1}{t^2 + 1} \).

Upon combining, you notice that these derivatives cancel each other out, demonstrating how meticulous differentiation leads us to the discovery that the derivative of this particular function is zero.