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Magazine Chapter 12: Problem 7
Calculate \(\frac{d^{2} y}{d x^{2}}\). \(y=e^{-(x-1)}-x\)
Short Answer
Expert verified
The second derivative of the given function \(y = e^{-(x-1)} - x\) is \(\frac{d^2 y}{d x^2} = e^{-(x-1)}\).
Step by step solution
01
Find the first derivative of the function
To differentiate the given function, we use the sum rule of differentiation where the derivative of a sum or difference is the sum or difference of the derivatives. Differentiate each term separately with respect to x.
Given function:
\(y = e^{-(x-1)} - x\)
First term derivative: Since the first term is an exponential function, we use the chain rule of differentiation. The chain rule states that if a function is a composite of two functions, the derivative of the composite function is the derivative of the outer function times the derivative of the inner function.
Let the outer function be \(f(u) = e^{-u}\) and inner function be \(u = x-1\). Find the derivative of each function and apply the product rule.
\[f'(u) = -e^{-u}\] and \[u'(x) = 1\]
Applying the chain rule,
\(\frac{d}{dx}(e^{-(x-1)}) = -e^{-(x-1)}\)
Second term derivative: The second term is a linear function, which is easy to differentiate:
\(\frac{d}{dx}(-x) = -1\)
Now, combining both derivatives:
\(\frac{dy}{dx} = -e^{-(x-1)} - 1\)
02
Find the second derivative of the function
Now we need to find the second derivative of the given function. As in the previous step, use the sum rule of differentiation:
Given first derivative function:
\(\frac{dy}{dx} = -e^{-(x-1)} - 1\)
First term derivative: Again, apply the chain rule to the exponential function.
Outer function: \(f(u) = -e^{-u}\) and inner function: \(u = x-1\)
Find the derivative of each function.
\[f'(u) = e^{-u}\] and \[u'(x) = 1\]
Applying the chain rule,
\(\frac{d^2}{dx^2}(-e^{-(x-1)}) = e^{-(x-1)}\)
Second term derivative: The second term is a constant, and the derivative of a constant is 0.
\(\frac{d^2}{dx^2}(-1) = 0\)
Now, combining both derivatives:
\(\frac{d^2 y}{d x^2} = e^{-(x-1)}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a fundamental concept in calculus, focusing on how a function changes. It's all about finding the derivative of a function, representing the rate at which the function's value changes as its input changes. To put this simply, if you know something about how a position changes over time, you can find velocity; if you know velocity, you can find acceleration. That's what differentiation does鈥攁 tool to explore these relationships.
In the process of differentiation, we often use rules like the sum and difference rules. These rules allow us to break down complex expressions into simpler parts. For a function like \( y = e^{-(x-1)} - x \), we can separately find the derivatives of each term and then combine them. This approach makes our work easier, ensuring accuracy and clarity.
One key aspect is recognizing when a function includes nested expressions, where we'll need more specialized techniques like the chain rule.
In the process of differentiation, we often use rules like the sum and difference rules. These rules allow us to break down complex expressions into simpler parts. For a function like \( y = e^{-(x-1)} - x \), we can separately find the derivatives of each term and then combine them. This approach makes our work easier, ensuring accuracy and clarity.
One key aspect is recognizing when a function includes nested expressions, where we'll need more specialized techniques like the chain rule.
Chain Rule
The chain rule is an essential technique in differentiation, especially when dealing with composite functions. Composite functions are like a nest of functions鈥攐ne inside another鈥攔equiring a unique approach to differentiate.
Consider a function \( y = e^{-(x-1)} \). Here, we have an exponential function wrapped around an inner function, \( u = x-1 \), indicating that the chain rule is necessary.
The chain rule can be simplified as a two-step process:
Practicing with the chain rule helps tackle more complex problems, building a deeper understanding of how changes in variables affect wrapped functions.
Consider a function \( y = e^{-(x-1)} \). Here, we have an exponential function wrapped around an inner function, \( u = x-1 \), indicating that the chain rule is necessary.
The chain rule can be simplified as a two-step process:
- First, find the derivative of the outer function regarding its inner counterpart.
- Then, multiply that result by the derivative of the inner function.
Practicing with the chain rule helps tackle more complex problems, building a deeper understanding of how changes in variables affect wrapped functions.
Exponential Functions
Exponential functions introduce unique characteristics in differentiable expressions. Unlike polynomial functions, exponential functions grow at a rate proportional to their current value, leading to distinctive behaviors.
The function \( y = e^{-(x-1)} \) is an example where differentiation requires special considerations. Exponential functions generally have the property that their derivative is a constant multiple of themselves. Specifically for \( e^u \), the derivative is as straightforward as \( e^u \), and when you factor in a negative exponent like \( -u \), it adjusts to \( -e^{u} \), requiring careful attention to signs.
This quality makes exponential functions remarkably consistent and predictable. For instance, in differentiating \( e^{-(x-1)} \), even when you move to the second derivative, it maintains its basic structural form, albeit with adjustments from the chain rule.
The function \( y = e^{-(x-1)} \) is an example where differentiation requires special considerations. Exponential functions generally have the property that their derivative is a constant multiple of themselves. Specifically for \( e^u \), the derivative is as straightforward as \( e^u \), and when you factor in a negative exponent like \( -u \), it adjusts to \( -e^{u} \), requiring careful attention to signs.
This quality makes exponential functions remarkably consistent and predictable. For instance, in differentiating \( e^{-(x-1)} \), even when you move to the second derivative, it maintains its basic structural form, albeit with adjustments from the chain rule.
Second Derivative
The second derivative is a deeper dive into how a function's rate of change itself changes over time. It's literally the differentiation of a derivative and represents the acceleration of an initial function, much like how acceleration is the rate of change of velocity.
When we calculate the first derivative of \( y = e^{-(x-1)} - x \), we get \( -e^{-(x-1)} - 1 \). For the second derivative, we focus on differentiating this result:
Understanding second derivatives aids in analyzing concavity, acceleration, and the overall behavior of functions beyond just their slopes.
When we calculate the first derivative of \( y = e^{-(x-1)} - x \), we get \( -e^{-(x-1)} - 1 \). For the second derivative, we focus on differentiating this result:
- The exponential component \( -e^{-(x-1)} \) transforms, through the chain rule, into \( e^{-(x-1)} \) without the negative sign.
- The constant \( -1 \) vanishes, as constants differentiate to zero.
Understanding second derivatives aids in analyzing concavity, acceleration, and the overall behavior of functions beyond just their slopes.
