Problem 55 - Calculus Early Transcendentals

Question

Find the indicated derivative. $$ y=\cot ^{3}(\pi-\theta) ; \text { find } \frac{d y}{d \theta} $$

Step-by-Step Solution

Answer

\( \frac{dy}{d\theta} = 3\cot^2(\pi - \theta) \csc^2(\pi - \theta) \).

Step 1: Identify the Chain Rule

To find the derivative of a composite function, we use the chain rule. Here, we have a composition of functions: the inner function is \( u = \pi - \theta \) and the outer function is \( y = \cot^3(u) \).

Step 2: Differentiate the Outer Function

First, differentiate the outer function \( y = \cot^3(u) \) with respect to \( u \), treating \( u \) as a variable. \( \frac{d}{du}[\cot^3(u)] = 3\cot^2(u) \cdot (-\csc^2(u)) \).This uses the chain rule where \( d[\cot(u)]/du = -\csc^2(u) \). Thus, \( \frac{d}{du}[\cot^3(u)] = -3\cot^2(u) \csc^2(u) \).

Step 3: Differentiate the Inner Function

Next, find the derivative of the inner function \( u = \pi - \theta \) with respect to \( \theta \).\( \frac{du}{d\theta} = \frac{d}{d\theta}[\pi - \theta] = -1 \).

Step 4: Apply the Chain Rule

Combine the results from the previous steps by multiplying the derivative of the outer function by the derivative of the inner function.\( \frac{dy}{d\theta} = \frac{d}{du}[\cot^3(u)] \cdot \frac{du}{d\theta} = -3\cot^2(u) \csc^2(u) \cdot (-1) \).This simplifies to:\( \frac{dy}{d\theta} = 3\cot^2(\pi - \theta) \csc^2(\pi - \theta) \).

Step 5: Simplify and Report the Final Answer

The simplified derivative is:\( \frac{dy}{d\theta} = 3\cot^2(\pi - \theta) \csc^2(\pi - \theta) \).There's no further simplification needed, so this is our final result.

Key Concepts

Chain Rule

The Chain Rule is a cornerstone concept in calculus, often used when finding derivatives of composite functions. Imagine you're dealing with a function inside another function, kind of like a box within a box. The outer box is altered by the inner box, meaning you must handle each separately. In our exercise, we have a function where - The inner function: \( u = \pi - \theta \) - The outer function: \( y = \cot^3(u) \) By employing the Chain Rule, we find derivatives step-by-step: 1. **Differentiate the outer function**: Treat the inner function as if it were a temporary variable. Thus, for the function \( \cot^3(u) \), find its derivative with respect to \( u \). 2. **Differentiate the inner function**: Derive \( u \) concerning \( \theta \).3. **Combine the derivatives**: Finally, multiply the derivative of the outer function by the derivative of the inner function to obtain the final result. Using this structured method makes it easier to decipher complex derivative problems, letting us solve them piece by piece.

Composite Function

A Composite Function arises when one function is nested inside another. This is common in calculus problems and necessitates special rules for differentiation. In our exercise, we have two functions combined:- **Inner Function**: \( u = \pi - \theta \)- **Outer Function**: \( y = \cot^3(u) \)This nesting means that the value of one function impacts how the second function behaves. To solve such derivatives:- Look for an inner and outer function relationship- Use the Chain Rule to handle them separatelyHaving a composite function requires us to adopt a methodical approach. Each layer or function part has to be peeled back individually and worked through. This clarity is essential for multitiered functions, enabling smarter and more efficient problem-solving in calculus.

Trigonometric Derivatives

Trigonometric Derivatives are a set of rules applied specifically to derivatives involving trigonometric functions. These functions, like sine, cosine, and cotangent, have their unique derivatives. Recognizing these is fundamental in calculus.In our specific problem, we dealt with the - **Derivative of Cotangent**: which is \( \frac{d}{du} [\cot(u)] = -\csc^2(u) \)When differentiating \( y = \cot^3(u) \), we used the chain rule where we acknowledged these rules for differentiation. The outer derivative involves the exponent rule combined with trigonometric properties:\[\frac{d}{du}[\cot^3(u)] = 3\cot^2(u) \cdot ( -\csc^2(u) )\]Understanding how these derivatives function and combine is pivotal. It allows for solving problems that incorporate the intricacies of both algebraic operations and trigonometric behaviors in calculus.