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Find the derivative of the following functions. $$f(x)=x e^{-x}$$

Short Answer

Expert verified
Answer: The derivative of the function $$f(x) = x e^{-x}$$ is $$f'(x) = e^{-x} - x e^{-x}$$.

Step by step solution

01

Find the derivative of f(x)

To find the derivative of $$f(x) = x$$, we will apply the power rule, which states that if $$f(x) = x^n$$, then $$f'(x) = nx^{n-1}$$. In this case, we have $$n=1$$, so the derivative of $$f(x)$$ is: $$f'(x) = 1x^{1-1}=1x^0=1$$.
02

Find the derivative of g(x)

To find the derivative of $$g(x) = e^{-x}$$, we will apply the chain rule, which states that if $$g(x) = h(u(x))$$, then $$g'(x) = h'(u(x))u'(x)$$. In this case, we have $$h(u) = e^u$$ and $$u(x) = -x$$, so the chain rule gives $$g'(x) = e^{-x} (-1) = -e^{-x}$$.
03

Apply the product rule

Now we will apply the product rule to find the derivative of our given function $$f(x)=x e^{-x}$$. As we've found earlier: $$f'(x) = 1$$ and $$g'(x) = -e^{-x}$$, so using the product rule: $$h'(x) = f'(x)g(x) + f(x)g'(x)= (1)(e^{-x}) + (x)(-e^{-x}) = e^{-x} - x e^{-x}$$ So, the derivative of the given function is: $$f'(x) = e^{-x} - x e^{-x}$$.

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Key Concepts

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

Power Rule
The power rule is a fundamental technique in calculus for finding derivatives of polynomial functions. It is especially useful when dealing with functions of the form \( f(x) = x^n \), where \( n \) is any real number. The rule states that the derivative \( f'(x) \) is found by multiplying \( n \) by \( x \) raised to the power of \( n-1 \).
For instance, if \( f(x) = x^2 \), applying the power rule gives \( f'(x) = 2x^{2-1} = 2x \).
This simplifies the process of finding derivatives significantly, especially as functions grow more complex.

Applying Power Rule:
  • Identify the power \( n \) of \( x \).
  • Multiply the expression by the power \( n \).
  • Decrease the power by 1.
  • Write the simplified expression as the derivative.
It is crucial to use the power rule correctly when it's applicable, as it forms the basis for more advanced derivative techniques.
Chain Rule
The chain rule is a powerful tool in calculus, used to differentiate composite functions. Composite functions are those where one function is nested within another, such as \( h(x) = f(g(x)) \). To differentiate such functions, the chain rule provides a systematic way to do so.
The essence of the chain rule can be captured with the equation \( h'(x) = f'(g(x))g'(x) \).

Steps to Using the Chain Rule:
  • Differentiate the outer function while keeping the inner function intact.
  • Multiply by the derivative of the inner function.
Let's say \( g(x) = e^{-x} \). Here \( h(u) = e^u \) where \( u(x) = -x \). The outer function \( e^u \) differentiates to \( e^u \) and the derivative of \( u(x) \) would be \( -1 \), thus leading to \( -e^{-x} \) for \( g'(x) \).
The chain rule is critical as it allows us to tackle more complicated functions by breaking them into manageable parts.
Product Rule
When faced with differentiating products of functions, the product rule becomes indispensable. If you have two functions, \( f(x) \) and \( g(x) \), that are multiplied together, the product rule provides a method to find the derivative of their product directly.
The formula for the product rule is \( (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \).

Applying the Product Rule:
  • Find the derivative of the first function \( f'(x) \).
  • Multiply it by the second function \( g(x) \).
  • Find the derivative of the second function \( g'(x) \).
  • Multiply it by the first function \( f(x) \).
  • Add these two products together to get the final derivative.
In our exercise, multiplying the derivative of \( x \) which is 1, by \( e^{-x} \), and adding \( x \) times the derivative of \( e^{-x} \) which is \(-e^{-x}\) provides the full derivative: \( e^{-x} - x e^{-x} \).
The product rule complements the power and chain rules, providing a complete toolkit for differentiating a wide range of functions.

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Most popular questions from this chapter

a. Identify the inner function \(g\) and the outer function \(f\) for the composition \(f(g(x))=e^{k x},\) where \(k\) is a real number. b. Use the Chain Rule to show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x}\).

A thin copper rod, 4 meters in length, is heated at its midpoint, and the ends are held at a constant temperature of \(0^{\circ} .\) When the temperature reaches equilibrium, the temperature profile is given by \(T(x)=40 x(4-x),\) where \(0 \leq x \leq 4\) is the position along the rod. The heat flux at a point on the rod equals \(-k T^{\prime}(x),\) where \(k>0\) is a constant. If the heat flux is positive at a point, heat moves in the positive \(x\) -direction at that point, and if the heat flux is negative, heat moves in the negative \(x\) -direction. a. With \(k=1,\) what is the heat flux at \(x=1 ?\) At \(x=3 ?\) b. For what values of \(x\) is the heat flux negative? Positive? c. Explain the statement that heat flows out of the rod at its ends.

Let \(f(x)=\cos ^{2} x+\sin ^{2} x\). a. Use the Chain Rule to show that \(f^{\prime}(x)=0\). b. Assume that if \(f^{\prime}=0,\) then \(f\) is a constant function. Calculate \(f(0)\) and use it with part (a) to explain why \(\cos ^{2} x+\sin ^{2} x=1\).

Let \(C(x)\) represent the cost of producing \(x\) items and \(p(x)\) be the sale price per item if \(x\) items are sold. The profit \(P(x)\) of selling x items is \(P(x)=x p(x)-C(x)\) (revenue minus costs). The average profit per item when \(x\) items are sold is \(P(x) / x\) and the marginal profit is dP/dx. The marginal profit approximates the profit obtained by selling one more item given that \(x\) items have already been sold. Consider the following cost functions \(C\) and price functions \(p\). a. Find the profit function \(P\). b. Find the average profit function and marginal profit function. c. Find the average profit and marginal profit if \(x=a\) units are sold. d. Interpret the meaning of the values obtained in part \((c)\). $$\begin{aligned} &C(x)=-0.04 x^{2}+100 x+800, p(x)=200-0.1 x,\\\ &\bar{a}=1000 \end{aligned}$$

Let \(b\) represent the base diameter of a conifer tree and let \(h\) represent the height of the tree, where \(b\) is measured in centimeters and \(h\) is measured in meters. Assume the height is related to the base diameter by the function \(h=5.67+0.70 b+0.0067 b^{2}\). a. Graph the height function. b. Plot and interpret the meaning of \(\frac{d h}{d b}\).

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