Problem 9
Question
In Exercises \(7-12,\) find the indicated derivatives. $$ \frac{d s}{d t} \quad \text { if } \quad s=\frac{t}{2 t+1} $$
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{1}{(2t+1)^2} \).
1Step 1: Identify the function and its components
The given function is \( s = \frac{t}{2t+1} \). We need to take the derivative of \( s \) with respect to \( t \). The function is a rational function, where the numerator is \( t \) and the denominator is \( 2t + 1 \).
2Step 2: Recall the Quotient Rule
The quotient rule for derivatives states that if you have a function \( s = \frac{u}{v} \), then its derivative is given by \( \frac{d s}{d t} = \frac{v \frac{d u}{d t} - u \frac{d v}{d t}}{v^2} \). Here, \( u = t \) and \( v = 2t + 1 \).
3Step 3: Differentiate the numerator and denominator
Calculate the derivative of the numerator, \( u = t \), which is \( \frac{d u}{d t} = 1 \). Next, calculate the derivative of the denominator, \( v = 2t + 1 \), which is \( \frac{d v}{d t} = 2 \).
4Step 4: Apply the Quotient Rule
Using the quotient rule: \[ \frac{d s}{d t} = \frac{(2t + 1) \cdot 1 - t \cdot 2}{(2t + 1)^2} \]. Simplify the expression to obtain \( \frac{d s}{d t} = \frac{2t + 1 - 2t}{(2t + 1)^2} = \frac{1}{(2t + 1)^2} \).
Key Concepts
Quotient RuleRational FunctionsDifferentiation Techniques
Quotient Rule
The Quotient Rule is a fundamental differentiation technique used when you have a function that is the ratio of two other functions. In mathematics, if you have a function of the form \( s = \frac{u}{v} \), where both \( u \) and \( v \) are differentiable, the derivative of \( s \) with respect to \( t \) is given by:
For example, with the given function \( s = \frac{t}{2t+1} \), the numerator \( u = t \) and the denominator \( v = 2t+1 \). This technique allows us to systematically find the derivative even when the terms are complex.
One significant aspect of the Quotient Rule is its dependence on the proper calculation of derivatives for both parts of the fraction. Getting these parts correct will lead to an accurate derivative of the whole function.
- \( \frac{d s}{d t} = \frac{v \frac{d u}{d t} - u \frac{d v}{d t}}{v^2} \)
For example, with the given function \( s = \frac{t}{2t+1} \), the numerator \( u = t \) and the denominator \( v = 2t+1 \). This technique allows us to systematically find the derivative even when the terms are complex.
One significant aspect of the Quotient Rule is its dependence on the proper calculation of derivatives for both parts of the fraction. Getting these parts correct will lead to an accurate derivative of the whole function.
Rational Functions
Rational functions are expressions that embody the concept of division between two polynomial expressions. They take the form \( \frac{P(t)}{Q(t)} \), where \( P(t) \) and \( Q(t) \) are polynomials. In our exercise, the rational function is \( s = \frac{t}{2t+1} \).
These functions are pivotal in calculus as they often appear in mathematical models, statistics, and real-life optimization problems.
Grasping the dynamics of these functions gives you a strong base in understanding more sophisticated calculus concepts.
These functions are pivotal in calculus as they often appear in mathematical models, statistics, and real-life optimization problems.
- The behavior of rational functions, like their zeros and undefined points, is governed by both the numerator and the denominator.
- Understanding the domain, which is where the denominator is not zero, is crucial. For example, for \( s = \frac{t}{2t+1} \), the function would be undefined if \( 2t+1 = 0 \), thus, \( t eq -\frac{1}{2} \).
Grasping the dynamics of these functions gives you a strong base in understanding more sophisticated calculus concepts.
Differentiation Techniques
Differentiation in calculus is the process of finding the derivative of a function, which represents the rate of change of the function with respect to a variable. It's a key tool in understanding how functions behave and evolve.
There are several techniques for differentiating functions, but in this context, the focus is on:
Mastering these techniques gives students a comprehensive toolkit for analyzing and solving real-world problems they might encounter in sciences, engineering, and beyond.
There are several techniques for differentiating functions, but in this context, the focus is on:
- Chain Rule: Used when dealing with composite functions; you differentiate the outer function and multiply it by the derivative of the inner function.
- Quotient Rule: Specifically for finding the derivative of a function that is the ratio of two others, as discussed previously.
- Product Rule: When two functions are multiplied, this rule helps in finding their derivative by considering each part's contribution.
Mastering these techniques gives students a comprehensive toolkit for analyzing and solving real-world problems they might encounter in sciences, engineering, and beyond.
Other exercises in this chapter
Problem 9
Find the derivatives of the functions \(s=\frac{\sqrt{t}}{1+\sqrt{t}}\)
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In Exercises \(1-12,\) find \(d y / d x\) $$ y=\frac{4}{\cos x}+\frac{1}{\tan x} $$
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Heating a plate When a circular plate of metal is heated in an oven, its radius increases at the rate of 0.01 \(\mathrm{cm} / \mathrm{min}\) . At what rate is t
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Find \(d y / d x\) in Exercises \(1-10\) $$ y=x\left(x^{2}+1\right)^{-1 / 2} $$
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