Chapter 27: Problem 14
Find the derivatives of the given functions. $$u=\frac{e^{0.5 v}}{2 v}$$
Short Answer
Expert verified
The derivative of \( u = \frac{e^{0.5 v}}{2 v} \) is \( \frac{du}{dv} = \frac{e^{0.5v}(v - 2)}{4v^2} \).
Step by step solution
01
Understand the function
The given function is expressed as \[ u = \frac{e^{0.5v}}{2v} \]This is a quotient of the form \( \frac{f(v)}{g(v)} \) where \( f(v) = e^{0.5v} \) and \( g(v) = 2v \). To find the derivative, we will use the quotient rule.
02
Recall the quotient rule
The quotient rule states that for a function \( y = \frac{f(v)}{g(v)} \), the derivative \( \frac{dy}{dv} \) is given by\[ \frac{dy}{dv} = \frac{g(v)f'(v) - f(v)g'(v)}{(g(v))^2} \]
03
Differentiate the numerator
Differentiate the function \( f(v) = e^{0.5v} \) with respect to \( v \):\[ f'(v) = 0.5 e^{0.5v} \]This comes from the chain rule where the derivative of \( e^{kv} \) is \( ke^{kv} \), where \( k = 0.5 \).
04
Differentiate the denominator
The function \( g(v) = 2v \) is a simple linear function, so its derivative is\[ g'(v) = 2 \]
05
Substitute into the quotient rule
Substitute \( f(v) \), \( f'(v) \), \( g(v) \), and \( g'(v) \) into the quotient rule formula:\[ \frac{du}{dv} = \frac{2v(0.5e^{0.5v}) - e^{0.5v}(2)}{(2v)^2} \]
06
Simplify the expression
Simplify the expression:1. Calculate the numerator: \[ 2v(0.5e^{0.5v}) - e^{0.5v}(2) = v e^{0.5v} - 2e^{0.5v} \]2. Simplify the denominator: \[ (2v)^2 = 4v^2 \]Thus, the derivative is\[ \frac{du}{dv} = \frac{v e^{0.5v} - 2e^{0.5v}}{4v^2} \]Further simplification results in:\[ \frac{du}{dv} = \frac{e^{0.5v}(v - 2)}{4v^2} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The quotient rule is a powerful technique in calculus for finding the derivative of a quotient of two functions. When you have a function
- in the form \( y = \frac{f(v)}{g(v)} \),
- the derivative \( \frac{dy}{dv} \) is calculated using:
Chain Rule
The chain rule is another cornerstone of calculus, crucial for differentiating composite functions. In essence, the chain rule is applied when a function is nested within another function. For a function expressed as \[ f(g(x)) \]it tells us to differentiate the outer function and multiply it by the derivative of the inner function:\[\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\]This looks intricate at first glance, but let's break it down:
- The **outer function** is the main function which \( g(x) \) is inside. Differentiate this first as if the inner function wasn’t there.
- The **inner function** \( g(x) \) is then differentiated, representing its rate of change.
Exponential Functions
Exponential functions are fundamental in calculus, involving expressions of the form \[ e^x \]where \( e \) is the base of the natural logarithm, approximately equal to \( 2.71828 \). Key properties of exponential functions include:
- They have a constant rate of growth or decay, captured by \( e \).
- The derivative of \( e^x \) is unique because it is \( e^x \) itself—meaning the function grows at a rate proportional to its current value.
- It becomes \( ke^{kx} \), reflecting both the multiplicative factor and the inherent exponential growth.