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Magazine Chapter 4: Problem 62
\(61-62 .\) By calculating the first few derivatives, find a formula for the \(n\) th derivative of each function \((k\) is a constant \()\). $$ f(x)=e^{-k x} $$
Short Answer
Expert verified
The \( n \)-th derivative is \((-1)^n k^n e^{-kx}\).
Step by step solution
01
Understand the Function
The function given is \( f(x) = e^{-kx} \), where \( k \) is a constant. We need to find a formula for the \( n \)-th derivative of this function.
02
Calculate the First Derivative
The first derivative of \( f(x) = e^{-kx} \) can be found using the chain rule: \[ f'(x) = \frac{d}{dx}(e^{-kx}) = -k e^{-kx}. \]
03
Calculate the Second Derivative
Take the derivative of the first derivative:\[ f''(x) = \frac{d}{dx}(-k e^{-kx}) = (-k)(-k)e^{-kx} = k^2 e^{-kx}. \]
04
Calculate the Third Derivative
Take the derivative of the second derivative:\[ f'''(x) = \frac{d}{dx}(k^2 e^{-kx}) = (k^2)(-k)e^{-kx} = -k^3 e^{-kx}. \]
05
Identify the Pattern
From the derivatives calculated, we have:- First derivative: \( -k e^{-kx} \)- Second derivative: \( k^2 e^{-kx} \)- Third derivative: \( -k^3 e^{-kx} \)The pattern is that the \( n \)-th derivative is \((-1)^n k^n e^{-kx}\).
06
Write the Formula for the nth Derivative
Based on the observed pattern, the formula for the \( n \)-th derivative of \( f(x) = e^{-kx} \) is:\[ f^{(n)}(x) = (-1)^n k^n e^{-kx}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
nth derivative
Finding the \( n \)-th derivative of a function involves determining the derivative of a function enough times until you spot a pattern and can generalize it. For the function \( f(x) = e^{-kx} \), this process begins by calculating the initial derivatives and observing trends in the coefficients and sign changes.
- The first derivative \( f'(x) = -k e^{-kx} \) incorporates both the power of \( k \) and a change in sign.
- Each subsequent derivative multiplies by an additional \((-k)\), causing the sign to alternate and the power of \( k \) to increase efficiently.
chain rule
The chain rule is crucial when differentiating composite functions. It enables us to find the derivative of a function inside another function. For \( f(x) = e^{-kx} \), you identify \(-kx\) as the inner function and \( e^u \) (where \( u = -kx \)) as the outer function.
- The derivative of the outer function is simply \( e^u \).
- The derivative of the inner function \(-kx\) is \(-k\).
exponential function
Exponential functions have the characteristic form \( e^{u(x)} \), often showcasing the base \( e \), which is approximately 2.71828. The intriguing part of exponential functions is that their derivatives are in the same functional form as the original function. For \( f(x) = e^{-kx} \), the exponential function involves an exponent \(-kx\). Key properties:
- The exponential function's derivative remains an exponential function, \[ \frac{d}{dx}(e^{-kx}) = -k e^{-kx}. \]
- The formula \( f^{(n)}(x) = (-1)^n k^n e^{-kx} \) after several differentiations shows how exponentiation, particularly with a negative exponent, affects the derivative pattern.
pattern recognition
Recognizing patterns in sequences of derivatives is a vital skill in calculus differentiation. Initially, each derivative calculation might seem intricate, but with repeated computation comes a clearer picture of the derivative's structure. For \( f(x) = e^{-kx} \):
- First, focus on signs: the first derivative \( -k e^{-kx} \) establishes a sign that alternates with each differentiation.
- Next, observe powers: each derivative heightens the power of \( k \) sequentially, i.e., \( k, k^2, k^3, \ldots \).
