Chapter 23: Problem 31
Find the derivative of each of the given functions. $$f(R)=\sqrt{\frac{2 R+1}{4 R+1}}$$
Short Answer
Expert verified
The derivative of \( f(R) \) is \( f'(R) = \frac{-\sqrt{4R+1}}{(4R+1)^{2} \sqrt{2R+1}} \).
Step by step solution
01
Rewrite the Function for Easier Differentiation
The given function is \( f(R) = \sqrt{\frac{2R + 1}{4R + 1}} \). To make differentiation easier, rewrite the function as \( f(R) = \left(\frac{2R+1}{4R+1}\right)^{1/2} \). This way, we can use the chain rule and the quotient rule more effectively.
02
Apply the Chain Rule
To differentiate \( f(R) = \left( \frac{2R+1}{4R+1} \right)^{1/2} \), start by applying the chain rule, which states that if you have \( u(x)^{n} \), then the derivative is \( n \cdot u(x)^{n-1} \cdot u'(x) \). Here, \( u(R) = \left( \frac{2R+1}{4R+1} \right) \), and \( n = 1/2 \). Differentiate \( u(R)^{1/2} \) with respect to \( u(R) \): \[ \frac{d}{dR} \left( \frac{2R+1}{4R+1} \right)^{1/2} = \frac{1}{2} \left( \frac{2R+1}{4R+1} \right)^{-1/2} \cdot \frac{d}{dR}\left( \frac{2R+1}{4R+1} \right) \]
03
Differentiate Using the Quotient Rule
Now, apply the quotient rule to find \( u'(R) \), where \( u(R) = \frac{2R+1}{4R+1} \). The quotient rule states \( \left( \frac{v}{w} \right)' = \frac{v'w - vw'}{w^2} \).For \( v = 2R + 1 \) and \( w = 4R + 1 \), we have:- \( v' = 2 \),- \( w' = 4 \).Then:\[ \frac{d}{dR} \left( \frac{2R+1}{4R+1} \right) = \frac{(2)(4R+1) - (2R+1)(4)}{(4R+1)^2} = \frac{8R+2 - 8R-4}{(4R+1)^2} = \frac{-2}{(4R+1)^2} \]
04
Substitute and Simplify
Substitute \( u'(R) \) back into the expression from Step 2:\[ \frac{d}{dR} \left( \frac{2R+1}{4R+1} \right)^{1/2} = \frac{1}{2} \left( \frac{2R+1}{4R+1} \right)^{-1/2} \cdot \frac{-2}{(4R+1)^2} \]Simplify:- \( \frac{-1}{(4R+1)^2} \cdot \frac{1}{\sqrt{\frac{2R+1}{4R+1}}} \)- It becomes \( \frac{-1}{(4R+1)^2} \cdot \sqrt{\frac{4R+1}{2R+1}} \)Thus, the derivative is:\[ f'(R) = \frac{-\sqrt{4R+1}}{(4R+1)^{2} \sqrt{2R+1}} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
In calculus, finding the derivative of a function is a fundamental concept. The derivative measures how a function changes as its input changes. It is basically the slope of the function at any given point. This is crucial because it tells us how a function behaves and helps in solving problems related to rate of change, optimization, and more.
To differentiate a function like \( f(R) = \sqrt{\frac{2 R+1}{4 R+1}} \), one might need to rewrite the function in a more manageable form for applying derivative rules, such as the chain and quotient rules.
In our case, rewriting \( f(R) \) as \( \left( \frac{2R+1}{4R+1} \right)^{1/2} \) helps in applying these rules effectively. Learning to express functions in different ways is a key skill in calculus.
To differentiate a function like \( f(R) = \sqrt{\frac{2 R+1}{4 R+1}} \), one might need to rewrite the function in a more manageable form for applying derivative rules, such as the chain and quotient rules.
In our case, rewriting \( f(R) \) as \( \left( \frac{2R+1}{4R+1} \right)^{1/2} \) helps in applying these rules effectively. Learning to express functions in different ways is a key skill in calculus.
Chain Rule
The chain rule is a powerful tool in calculus used when differentiating composite functions. If you have a function that consists of another function, such as \( \, u(x)^n \, \), the derivative is calculated using the chain rule:
- The outer function is differentiated according to its rule, e.g., power rule which gives \( n \cdot u(x)^{n-1} \).
- The inner function \( u(x) \) is differentiated separately.
- Multiply the derivative of the outer function by the derivative of the inner function.
Quotient Rule
The quotient rule is essential for differentiating functions presented as a fraction of two other functions. It's used when you have a function in the form of \( \frac{v(x)}{w(x)} \). The rule states that the derivative is:
\[ \left( \frac{v}{w} \right)' = \frac{v'w - vw'}{w^2} \]
Where:
Understanding how to accurately apply the quotient rule helps in managing functions where two different rates of change are involved, making it a key component in calculus derivatives.
\[ \left( \frac{v}{w} \right)' = \frac{v'w - vw'}{w^2} \]
Where:
- \( v \) is the numerator function and \( w \) the denominator function.
- \( v' \) is the derivative of \( v \) and \( w' \) is the derivative of \( w \).
Understanding how to accurately apply the quotient rule helps in managing functions where two different rates of change are involved, making it a key component in calculus derivatives.