/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Find the differential of the fun... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 18

Find the differential of the function at the indicated number. $$ f(x)=2^{-x^{2}} ; \quad x=1 $$

Short Answer

Expert verified
The differential of the function \(f(x)=2^{-x^2}\) at \(x=1\) is \(\frac{df}{dx}(1) = -\ln(2)\).

Step by step solution

01

Find the derivative of the function

To find the derivative of the function \(f(x) = 2^{-x^2}\), we can use the chain rule. The chain rule states that \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \] In this case, let \(f(u) = 2^u\) and \(g(x) = -x^2\). Then we have \(f(g(x)) = 2^{-x^2}\). Now, we will find the derivative of \(f(u)\) and \(g(x)\). \[ f'(u) = \frac{d}{du}(2^u) = 2^u \ln(2) \] \[ g'(x) = \frac{d}{dx}(-x^2) = -2x \] Using the chain rule, we can now find the derivative of our function. \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) = 2^{-x^2} \ln(2) \cdot (-2x) \] Therefore, the derivative of the function is \(\frac{df}{dx} = 2^{-x^2} \ln(2)(-2x)\).
02

Evaluate the derivative at the given point

Now, we need to find the value of the derivative at \(x = 1\). We substitute \(x = 1\) into the derivative: \[ \frac{df}{dx}(1) = 2^{-1^2} \ln(2)(-2(1)) = 2^{-1} \ln(2)(-2) = \frac{1}{2} \ln(2)(-2) \] So the derivative of the function at the point \(x = 1\) is \(\frac{df}{dx}(1) = -\ln(2)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Chain Rule
Mastering the chain rule is essential for any student diving into differential calculus. Think of it as a way to peel back the layers of an onion. When you have a function that's nested within another function, like a Russian doll, the chain rule helps you differentiate these composite functions. It essentially states that if you have a function inside another function鈥攍et's say, f inside g鈥攜ou differentiate the outer function while keeping the inner one as is, and then multiply by the derivative of the inner function.

In our example, we had the function f(x) = 2^{-x^2}. Imagining that as an onion, the outermost layer is 2^u, and inside it, taking the place of u, is our second layer, -x^2. Using the chain rule, we differentiate the outer layer with respect to u, then we multiply that by the derivative of our inner layer, -x^2, with respect to x.
Derivative of Exponential Functions
Now, diving into exponential functions, these are the ones that look like a^x, where a is a constant and x is the exponent. When we differentiate these functions, a new component comes into play鈥攖he natural logarithm, denoted as ln.

The derivative of an exponential function a^x is a^x ln(a). Why is that? It's because the natural logarithm is the inverse of the exponential function with base e, the Euler's number. And in differentiation, the natural logarithm helps us adjust the rate of change when the base is not e. In the case of our function 2^u where u=-x^2, after applying this rule, we find that its derivative is 2^u ln(2), reflecting the relationship between exponentials and logarithms.
Evaluating Derivatives
Once we've calculated the derivative鈥攐r the rate at which our function changes鈥攚e might need to evaluate it at a particular point to understand the behavior of the function at that specific location. Evaluating derivatives is basically plugging in a value for x and calculating the result.

To evaluate the derivative at a given point, we just replace x with the number we're interested in. If our derived formula is f'(x) = 2^{-x^2}ln(2)(-2x) and we need to know the value at x=1, we substitute 1 for every x in the derivative. This allows us to determine the instantaneous rate of change of our function at x=1, effectively giving us a snapshot of the function's behavior at that very point. For deeper understanding, students should practice substituting different values into the derivative, which allows them to explore how the function changes at various points along its domain.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Percent of Females in the Labor Force Based on data from the U.S. Census Bureau, the following model giving the percent of the total female population in the civilian labor force, \(P(t)\), at the beginning of the \(t\) h decade \((t=0\) corresponds to the year 1900 ) was constructed. \(P(t)=\frac{74}{1+2.6 e^{-0.166 r+0.04536 x^{2}-0.0066 r^{3}}} \quad 0 \leq t \leq 11\) a. What was the percent of the total female population in the civilian labor force at the beginning of 2000 ? b. What was the growth rate of the percentage of the total female population in the civilian labor force at the beginning of 2000 ? Source: U.S. Census Bureau.

Two families of curves are orthogonal trajectories of each other if every curve of one family is orthogonal to every curve in the other family. In Exercises \(93-96\), (a) show that the given families of curves are orthogonal to each other, and (b) sketch a few members of each family on the same set of axes. $$ x^{2}+y^{2}=c^{2}, \quad y=k x, \quad c, k \text { constants } $$

Bimolecular Reaction In a bimolecular reaction \(A+B \rightarrow M, a\) moles per liter of \(A\) and \(b\) moles per liter of \(B\) are combined. The number of moles per liter that have reacted after time \(t\) is given by $$ x=\frac{a b\left[1-e^{(b-a) k t}\right]}{a-b e^{(b-a) k t}} \quad a \neq b $$ where the positive number \(k\) is called the velocity constant. Find an expression for \(x\) if \(a=b\), and find \(\lim _{t \rightarrow \infty} x\). Interpret your results.

Effect of an Earthquake on a Structure To study the effect an earthquake has on a structure, engineers look at the way a beam bends when subjected to an earth tremor. The equation $$ D=a-a \cos \left(\frac{\pi h}{2 L}\right) \quad 0 \leq h \leq L $$ where \(L\) is the length of a beam and \(a\) is the maximum deflection from the vertical, has been used by engineers to calculate the deflection \(D\) at a point on the beam \(h \mathrm{ft}\) from the ground. Suppose that a 10 -ft vertical beam has a maximum deflection of \(\frac{1}{2} \mathrm{ft}\) when subjected to an external force. Using differentials, estimate the difference in the deflection between the point midway on the beam and the point \(\frac{1}{10} \mathrm{ft}\) above it.

Find the derivative of the function. $$ y=\left(x^{2}+1\right) \tan ^{-1} x $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.