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Computing the derivative of \(f(x)=e^{-x}\) a. Use the definition of the derivative to show that \(\frac{d}{d x}\left(e^{-x}\right)=e^{-x} \cdot \lim _{h \rightarrow 0} \frac{e^{-h}-1}{h}\) b. Show that the limit in part (a) is equal to \(-1 .\) (Hint: Use the facts that \(\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1\) and \(e^{x}\) is continuous for all \(x\).) c. Use parts (a) and (b) to find the derivative of \(f(x)=e^{-x}\)

Short Answer

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

Step by step solution

01

Recall the definition of the derivative

The derivative of a function is defined as follows: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$
02

Apply the definition to the given function

Now, let's apply the definition of the derivative to the given function \(f(x) = e^{-x}\): $$ f'(x) = \lim_{h \to 0} \frac{e^{-(x+h)} - e^{-x}}{h} $$
03

Factor out \(e^{-x}\)

Now factor out the \(e^{-x}\) term from the numerator: $$ f'(x) = \lim_{h \to 0} \frac{e^{-x}(e^{-h} - 1)}{h} $$ $$ f'(x) = e^{-x} \cdot \lim_{h \to 0} \frac{e^{-h} - 1}{h} $$ #b. Proving the limit is equal to -1#
04

Use given hint to compute the limit

It is given that \(\lim_{h \to 0} \frac{e^{h} - 1}{h} = 1\). We need to find the limit of another expression: $$\lim_{h \to 0} \frac{e^{-h} - 1}{h}$$ Let \(u = -h\). Then, as \(h \to 0\), \(u \to 0\). So, we have: $$ \lim_{u \to 0} \frac{e^{u} - 1}{-u} = -1 \cdot \lim_{u \to 0} \frac{e^{u} - 1}{u} = -1 \cdot 1 = -1$$ #c. Finding the derivative of \(f(x) = e^{-x}\) using parts (a) and (b)#
05

Substitute the limit value in part (a)

Now, substitute the value of the limit we found in part (b) into the expression from part (a): $$ f'(x) = e^{-x} \cdot (-1) $$
06

Write the final derivative

Therefore, the derivative of the function \(f(x) = e^{-x}\) is: $$ f'(x) = -e^{-x} $$

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Key Concepts

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

Definition of the Derivative
The derivative of a function is a fundamental concept in calculus. It measures how a function changes as its input changes. Imagine slipping on a shoe. The derivative gives us the rate at which the shoe fits more or less snugly as the size changes. In mathematical terms, the derivative of a function \( f(x) \) at a point \( x \) is a limit:
  • \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
The expression \( f(x+h) - f(x) \) calculates the change in the function's value. The \( h \) part is a small change in the \( x \) value. Taking the limit as \( h \) approaches zero gives us the precise rate of change, or the derivative. This formula helps us see how a tiny shift in \( x \) affects \( f(x) \). When you apply this to an exponential function like \( f(x) = e^{-x} \), you enter its powerful world of calculus where even tiny changes have clear, computable effects.
Limit of a Function
Limits help us understand what happens to a function as it approaches a particular point or value. These are incredibly useful in calculus for calculating derivatives and integrals. Essentially, a limit evaluates the behavior of a function as its inputs get infinitely close to some point. To find such a limit for a derivative, you'd look at an expression like this:
  • \( \lim_{h \to 0} \frac{e^{-h} - 1}{h} \)
In our exercise, we used a handy hint: \( \lim_{h \to 0} \frac{e^{h} - 1}{h} = 1 \). Substituting back, with a change of variables, shows that the original limit equals \(-1\). This helps in confirming parts of the derivative calculation and helps us visualize the function's behavior near that point. Understanding limits is crucial because they form the backbone of calculating derivatives and integrals which describe rates of change.
Exponential Function Continuity
Exponential functions, such as \( e^x \), are continuous. Continuity means there are no breaks, jumps, or holes in the graph of the function. For a function to be continuous at a point \( x \), three things must be true:
  • \( f(x) \) is defined at the point.
  • \( \lim_{y \to x} f(y) \) exists, meaning it approaches a particular value as \( y \) gets close to \( x \).
  • \( \lim_{y \to x} f(y) = f(x) \), linking the value approached by the limit to the actual function value.
Exponential functions maintain this continuity over their entire domain. This property is essential when calculating limits and derivatives smoothly. Knowing that \( e^x \) is continuous allows us to substitute limits directly, ensuring correct derivative calculation without unexpected jumps or oddities. Embracing continuity simplifies analysis, especially with calculus operations.

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Most popular questions from this chapter

A study conducted at the University of New Mexico found that the mass \(m(t)\) (in grams) of a juvenile desert tortoise \(t\) days after a switch to a particular diet is described by the function \(m(t)=m_{0} e^{0.004 t},\) where \(m_{0}\) is the mass of the tortoise at the time of the diet switch. If \(m_{0}=64\) evaluate \(m^{\prime}(65)\) and interpret the meaning of this result. (Source: Physiological and Biochemical Zoology, 85,1,2012 )

The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots.\) c. Use the functions found in part (b) to graph the given equation. \(x^{4}=2\left(x^{2}-y^{2}\right)\) (eight curve)

The population of a culture of cells after \(t\) days is approximated by the function \(P(t)=\frac{1600}{1+7 e^{-0.02 t}},\) for \(t \geq 0\) a. Graph the population function. b. What is the average growth rate during the first 10 days? c. Looking at the graph, when does the growth rate appear to be a maximum? d. Differentiate the population function to determine the growth rate function \(P^{\prime}(t)\) e. Graph the growth rate. When is it a maximum and what is the population at the time that the growth rate is a maximum?

Tangency question It is easily verified that the graphs of \(y=x^{2}\) and \(y=e^{x}\) have no point of intersection (for \(x>0\) ), while the graphs of \(y=x^{3}\) and \(y=e^{x}\) have two points of intersection. It follows that for some real number \(2 < p < 3,\) the graphs of \(y=x^{p}\) and \(y=e^{x}\) have exactly one point of intersection (for \(x > 0) .\) Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

Carry out the following steps. a. Use implicit differentiation to find \(\frac{d y}{d x}\). b. Find the slope of the curve at the given point. $$x^{2 / 3}+y^{2 / 3}=2 ;(1,1)$$

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