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Chapter 3: Problem 21

Find the values of the derivatives. $$\left.\frac{d r}{d \theta}\right|_{\theta=0} \text { if } r=\frac{2}{\sqrt{4-\theta}}$$

Short Answer

Expert verified
\( \frac{1}{8} \)

Step by step solution

01

Differentiate the function

Begin by differentiating the function \( r = \frac{2}{\sqrt{4-\theta}} \) with respect to \( \theta \). Use the chain rule and power rule for differentiation. Rewrite the function as \( r = 2(4-\theta)^{-1/2} \).
02

Apply the Chain Rule

Differentiate \( r = 2 (4-\theta)^{-1/2} \) by first considering \( u = 4-\theta \), which implies \( rac{du}{d\theta} = -1 \). Now differentiate \( 2u^{-1/2} \) with respect to \( u \): \( \frac{d}{du}(2u^{-1/2}) = -u^{-3/2} \). Multiply this by \( \frac{du}{d\theta} \), giving \( \frac{dr}{d\theta} = 2(-1)\cdot(-u^{-3/2}) = u^{-3/2} \).
03

Substitute Back and Simplify

Substitute back \( u = 4-\theta \) into the derivative to get \( \frac{dr}{d\theta} = (4-\theta)^{-3/2} \).
04

Evaluate at \( \theta = 0 \)

Substitute \( \theta = 0 \) into the derivative expression: \( \left.\frac{dr}{d\theta}\right|_{\theta=0} = (4-0)^{-3/2} \). Simplify to get \((4)^{-3/2}\).
05

Calculate the Exact Value

Calculate \( (4)^{-3/2} \). Recognize that \( \sqrt{4} = 2 \), so \( 4^{-3/2} = (2^2)^{-3/2} = 2^{-3} = \frac{1}{8} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is a critical differentiation technique in calculus used when dealing with compositions of functions. When you have a function nested within another function, the chain rule allows you to differentiate them effectively. Here's how it works:
  • Identify the outer function and the inner function. In the case of our exercise, the outer function is a power rule problem \(u^{-1/2}\), while the inner function is \(4-\theta\).
  • First, differentiate the outer function with respect to the inner function. Here, we differentiated \(u^{-1/2}\) as \(-u^{-3/2}\).
  • Second, multiply this derivative by the derivative of the inner function with respect to the variable of interest, \(\theta\). In this example, \(\frac{du}{d\theta} = -1\).
By chaining these derivatives, the chain rule ensures that no part of the nested function is left undifferentiated, providing us with a complete derivative of the composed function.
Power Rule
The power rule simplifies differentiation when dealing with functions raised to a power. This rule states that if you have a function of the form \(x^n\), its derivative is \(nx^{n-1}\). It's a straightforward yet powerful tool in calculus.
Let's see how it was used in the exercise:
  • The expression \(2(4-\theta)^{-1/2}\) needed differentiation. Recognize it as having the form \(u^{-1/2}\).
  • Applying the power rule, \(u^{-1/2}\) becomes \(-1/2 \cdot u^{-3/2}\) after differentiation with respect to \(u\).
This process simplifies the calculation by reducing the power of the function in each step, making it easier to manipulate complex expressions. The correct use of the power rule is vital in reaching the solution correctly and efficiently.
Differentiation Techniques
Differentiation techniques include a variety of methods used in calculus to find the derivative of a function. Understanding these techniques is key for effectively tackling problems like the one in our exercise. Some common techniques include:
  • Chain Rule: As discussed, it's used for differentiating composite functions.
  • Power Rule: Utilized for functions involving powers of a variable.
  • Product Rule: This is applied when differentiating the product of two functions. Though not needed in our current example, it's useful to know for different problems.
  • Quotient Rule: Used to differentiate the division of two functions, which is another technique to keep in mind.
In our exercise, the chain rule and power rule were the main techniques used. Mastering these techniques requires practice, but they allow us to simplify the process of finding derivatives significantly. As you become familiar with these methods, solving calculus problems will become less challenging and more systematic.

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