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Chapter 4: Problem 60

\(59-60 .\) Find the second derivative of each function. $$ f(x)=e^{-x^{6} / 6} $$

Short Answer

Expert verified
The second derivative is \( f''(x) = (-5x^4 + x^{10}) e^{-x^6/6} \).

Step by step solution

01

Find the First Derivative

Given the function \( f(x) = e^{-x^6/6} \), we first need to find its first derivative \( f'(x) \). Use the chain rule for finding the derivative of the exponential function. Let \( u = -x^6/6 \), then \( f(x) = e^u \). The derivative of \( e^u \) with respect to \( x \) is \( e^u \cdot \frac{du}{dx} \). So, find \( \frac{du}{dx} = -x^5 \) using the power rule. Thus, \( f'(x) = e^{-x^6/6} \cdot (-x^5) = -x^5 e^{-x^6/6} \).
02

Find the Second Derivative

Now, we need to find the second derivative \( f''(x) \). Start by differentiating the first derivative \( f'(x) = -x^5 e^{-x^6/6} \) again using the product rule. Let \( u = -x^5 \) and \( v = e^{-x^6/6} \), where \( f'(x) = u \cdot v \). The product rule states \( (uv)' = u'v + uv' \). Differentiating \( u = -x^5 \) gives \( u' = -5x^4 \). For \( v = e^{-x^6/6} \), \( v' = e^{-x^6/6} \cdot (-x^5) \). Applying the product rule: \[ f''(x) = (-5x^4) e^{-x^6/6} + (-x^5) (-x^5 e^{-x^6/6}) = (-5x^4 + x^{10}) e^{-x^6/6} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Function Derivative
The derivative of an exponential function is a vital part of calculus. When dealing with exponential functions like \( f(x) = e^{u}\), the derivative can be found using a simple rule. The function \( e^{u} \) is unique because its derivative is itself, \( e^{u} \), times the derivative of the exponent \( u \).
This rule can be stated as follows:
  • Identify \( u \), the exponent in the exponential function.
  • Compute \( \frac{du}{dx} \), the derivative of \( u \) with respect to \( x \).
  • The derivative of \( e^{u} \) is \( e^{u} \cdot \frac{du}{dx} \).
In our exercise, we identified \( u = -\frac{x^{6}}{6} \). The key was to differentiate this part, using the power rule, resulting in an expression that multiplied with \( e^{u} \). This illustrates the intertwined use of different derivative rules in calculus.
Chain Rule
The chain rule is a powerful tool for finding the derivative of composite functions. Whenever you differentiate a function within another function, it's time to use the chain rule.
To use the chain rule:
  • First, determine the outer function and the inner function. In our case, \( f(x) = e^{-x^6/6} \), where \( e^u \) is the outer function and \( -\frac{x^6}{6} \) is the inner function \( u \).
  • Differentiate the outer function with respect to \( u \) and the inner function \( u \) with respect to \( x \).
  • Multiply the two derivatives to get the derivative of the original function.
In this exercise, we applied the chain rule to the exponential function. The derivative of the outer function, \( e^u \), is \( e^u \) itself, and this was multiplied by the derivative of \( u \) from the inner function, leading us to the right solution.
Power Rule
The power rule is a foundational principle in calculus when differentiating expressions with exponents. This rule is straightforward and highly effective for finding derivatives quickly.
To apply the power rule, simply:
  • If you have \( x^n \), multiply by the exponent \( n \), and reduce the power by one.
  • This yields \( \frac{d}{dx}x^n = n \cdot x^{n-1} \).
In our example problem, the power rule helped find the derivative of \( -x^6/6 \) by focusing only on \( x^6 \). We multiplied by 6 (our exponent) to get \( 6x^5 \) and then applied the factor from the division. This power rule step was pivotal for expressing the first derivative accurately.
Product Rule
The product rule is essential for differentiating expressions where two functions are multiplied together. It helps us when the derivative involves a product of two functions, say \( u(x) \) and \( v(x) \).
According to the product rule:
  • If \( y = uv \), then the derivative is \( y' = u'v + uv' \).
  • First, find the individual derivatives, \( u' \) and \( v' \).
  • Use these derivatives to apply the formula.
For our problem, first, the derivative was broken into a product of \( -x^5 \) and \( e^{-x^6/6} \). By differentiating each term separately and then using the product rule, we found \( f''(x) \). The product rule ensures that multiplying functions work seamlessly even within the context of differentiation.

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Most popular questions from this chapter

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