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Magazine Chapter 4: Problem 42
Find the derivative of each function. $$ f(z)=\frac{12}{1+2 e^{-z}} $$
Short Answer
Expert verified
The derivative is \( \frac{24e^{-z}}{(1+2e^{-z})^2} \).
Step by step solution
01
Identify the Function Type
The given function is \( f(z) = \frac{12}{1+2e^{-z}} \). This is a rational function because it can be expressed in the form of a numerator divided by a denominator. In this case, the numerator is 12, and the denominator is \(1 + 2e^{-z}\).
02
Apply the Quotient Rule
The quotient rule is used to differentiate functions of the form \( \frac{u}{v} \), and is given by the formula: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). Here, \( u = 12 \) and \( v = 1 + 2e^{-z} \).
03
Differentiate the Numerator
Differentiate the constant numerator \( u = 12 \), resulting in \( u' = 0 \), since the derivative of a constant is zero.
04
Differentiate the Denominator
The denominator \( v = 1 + 2e^{-z} \) can be differentiated using the chain rule. The derivative of \( 1 \) is zero and the derivative of \( 2e^{-z} \) is \( -2e^{-z} \), so \( v' = -2e^{-z} \).
05
Substitute into the Quotient Rule Formula
Substitute the derivatives \( u' = 0 \) and \( v' = -2e^{-z} \) into the quotient rule formula: \( \left(\frac{12}{1 + 2e^{-z}}\right)' = \frac{0 \cdot (1+2e^{-z}) - 12(-2e^{-z})}{(1+2e^{-z})^2} \).
06
Simplify the Expression
Simplify the expression: \( \left(\frac{12}{1 + 2e^{-z}}\right)' = \frac{24e^{-z}}{(1+2e^{-z})^2} \). This is the derivative of the given function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Functions
Rational functions are a specific type of function that involve fractions where both the numerator and denominator are polynomials. In simpler terms, it's when you have one polynomial divided by another.
For example, the function given in the exercise is \( f(z) = \frac{12}{1 + 2e^{-z}} \). Here, the numerator is the constant \(12\), and the denominator is \(1 + 2e^{-z}\). Although \(2e^{-z}\) involves an exponential function, it can still be considered within the rational form when expressed in terms of polynomials.
For example, the function given in the exercise is \( f(z) = \frac{12}{1 + 2e^{-z}} \). Here, the numerator is the constant \(12\), and the denominator is \(1 + 2e^{-z}\). Although \(2e^{-z}\) involves an exponential function, it can still be considered within the rational form when expressed in terms of polynomials.
- **Numerator (top part of the fraction):** It is usually a polynomial, but constants like \(12\) are also considered as polynomials of degree 0.
- **Denominator (bottom part of the fraction):** Usually a polynomial as well, which in this case involves an exponential expression incorporated within it.
Quotient Rule
The quotient rule is a method for finding the derivative of a function that is the quotient of two other functions. It is essential when working with rational functions.
The formula for the quotient rule is \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), where \(u\) is the numerator and \(v\) is the denominator. This formula helps us compute the derivative in a structured manner:
The formula for the quotient rule is \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), where \(u\) is the numerator and \(v\) is the denominator. This formula helps us compute the derivative in a structured manner:
- **Differentiate the Numerator \(u\):** First, find \(u'\), the derivative of the numerator.
- **Differentiate the Denominator \(v\):** Next, calculate \(v'\), the derivative of the denominator.
- **Apply the Quotient Rule Formula:** Substitute \(u'\), \(v\), \(u\), and \(v'\) into the formula to get the derivative.
Chain Rule
The chain rule is a fundamental tool used for differentiating composite functions. When a function is nested inside another, the chain rule is applied: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). This means you differentiate the outer function first and then multiply by the derivative of the inner function.
In the context of the provided exercise, the chain rule appears as part of differentiating the denominator \(v = 1 + 2e^{-z}\). Here, \(g(z) = -z\) is the inside function of the composite part \(2e^{-z}\):
In the context of the provided exercise, the chain rule appears as part of differentiating the denominator \(v = 1 + 2e^{-z}\). Here, \(g(z) = -z\) is the inside function of the composite part \(2e^{-z}\):
- **Inside Function \(g(z)\):** Derivative is \(-1\) since \((z)' = 1\).
- **Outside Function dereivative \(f(g(z)) = e^{g(z)}\):** Differentiated to give \(-2e^{-z}\). This results from applying the chain rule and multiplying by \(-1\).
