Problem 38 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ q=\cot \left(\frac{\sin t}{t}\right) $$

Step-by-Step Solution

Answer

The derivative is \(-\csc^2\left(\frac{\sin t}{t}\right) \cdot \frac{t \cos t - \sin t}{t^2}.\)

Step 1: Identify the composition of functions

The function given is a composition where \( q = \cot(u) \) and \( u = \frac{\sin t}{t} \). We will need to use the chain rule to differentiate it because it is the composition of two functions.

Step 2: Differentiate the outer function

Use the derivative of \( \cot(u) \) which is \( -\csc^2(u) \). Applying this to our function gives \( \frac{dq}{du} = -\csc^2\left(\frac{\sin t}{t}\right) \).

Step 3: Differentiate the inner function

The inner function is \( u = \frac{\sin t}{t} \). To differentiate this, use the quotient rule, \( \frac{d}{dt}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2} \), where \( f(t) = \sin t \) and \( g(t) = t \).

Step 4: Apply the quotient rule

Calculate \( f'(t) = \cos t \) and \( g'(t) = 1 \). Substitute these into the quotient rule to get:\[ \frac{du}{dt} = \frac{t(\cos t) - \sin t \cdot 1}{t^2} = \frac{t \cos t - \sin t}{t^2}. \]

Step 5: Apply the chain rule to find \( \frac{dq}{dt} \)

Use the chain rule \( \frac{dq}{dt} = \frac{dq}{du} \cdot \frac{du}{dt} \). Substitute the derivatives we found:\[ \frac{dq}{dt} = -\csc^2\left(\frac{\sin t}{t}\right) \cdot \frac{t \cos t - \sin t}{t^2}. \]

Key Concepts

Chain Rule

The Chain Rule is a fundamental principle in calculus that allows you to differentiate the composite of two or more functions. When you have a function nested within another function, the Chain Rule is your go-to technique. It helps us differentiate a composition of functions, effectively simplifying what could otherwise be a complex problem.

To visualize the Chain Rule, consider a function defined as \(y = f(g(x))\). In simple terms, rather than directly differentiating \(y\) with respect to \(x\), you perform differentiation in two stages:

• First, differentiate the outer function \(f\) with respect to the inner function \(g\), giving you \(f'(g(x))\).
• Then, differentiate the inner function \(g\) with respect to \(x\), resulting in \(g'(x)\).

The Chain Rule combines these derivatives to get \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). This method efficiently handles nested differentiation tasks by breaking them into manageable parts.

Quotient Rule

The Quotient Rule offers a way to differentiate functions that are ratios, meaning one function divided by another. This rule is vital when dealing with expressions in the form \(\frac{f(x)}{g(x)}\), enabling a systematic approach to find the derivative.

Applying the Quotient Rule involves the following steps:

• Identify the numerator function \(f(x)\) and the denominator function \(g(x)\).
• Differentiate both \(f(x)\) and \(g(x)\) to find \(f'(x)\) and \(g'(x)\).

Place these into the Quotient Rule formula: \[\frac{d}{dx}\left( \frac{f(x)}{g(x)} \right) = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}.\]This adjustment accounts for changes in both the numerator and denominator, ensuring the overall change is correctly represented. In our solution, the Quotient Rule was implements to differentiate \(u = \frac{\sin t}{t}\), which formed the inner part of our composite function relation.

Trigonometric Derivatives

Trigonometric derivatives are the derivatives of trigonometric functions, which are crucial in solving calculus problems involving these functions. These derivatives include functions like sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)), as well as their reciprocal functions like cosecant (\(\csc\)), secant (\(\sec\)), and cotangent (\(\cot\)).

Understanding and recalling the derivatives of these basic trigonometric functions is key:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).
• The derivative of \(\tan(x)\) is \(\sec^2(x)\).
• The derivative of \(\csc(x)\) is \(-\csc(x) \cot(x)\).
• The derivative of \(\sec(x)\) is \(\sec(x) \tan(x)\).
• The derivative of \(\cot(x)\) is \(-\csc^2(x)\).

In the solution, the derivative of \(\cot(u)\), which is \(-\csc^2(u)\), was used to differentiate the outer function in the chain, emphasizing the application of trigonometric derivatives in complex compositions.