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Chapter 3: Problem 18

Use Version I of the Chain Rule to calculate \(\frac{d y}{d x}\). $$y=e^{-x^{2}}$$

Short Answer

Expert verified
Question: Find the derivative of the function \(y = e^{-x^2}\) with respect to x. Answer: The derivative of the function \(y = e^{-x^2}\) with respect to x is \(\frac{dy}{dx} = -2xe^{-x^2}\).

Step by step solution

01

Differentiate the outer function y(u)

To differentiate \(y(u) = e^u\) with respect to u, we'll use the fact that the derivative of \(e^u\) is itself. So we have: $$\frac{dy}{du} = e^u$$
02

Differentiate the inner function u(x)

To differentiate \(u(x) = -x^2\) with respect to x, we'll use the power rule which states that the derivative of \(x^n\) is \(nx^{n-1}\). So we have: $$\frac{du}{dx} = -2x$$
03

Apply the Chain Rule

Now that we have the derivatives of both our inner and outer functions, we can use the Chain Rule to find the derivative of y with respect to x. The Chain Rule states: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ So, substituting the derivatives obtained earlier, we get: $$\frac{dy}{dx} = e^u \cdot (-2x)$$
04

Substitute back the inner function

Now that we have the derivative in terms of x and u, we need to substitute back the inner function, i.e., \(u(x) = -x^2\). So, we get: $$\frac{dy}{dx} = e^{-x^2} \cdot (-2x)$$ So, the derivative \(\frac{dy}{dx}\) of the given function \(y = e^{-x^2}\) is: $$\frac{dy}{dx} = -2xe^{-x^2}$$

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Key Concepts

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

Derivative
In calculus, a derivative represents how a function changes as its input changes. Think of it as a way to measure the rate at which things transform. In the context of a graph, the derivative gives the slope of the tangent line at any given point. This is crucial because it tells us how steep the graph is at that exact spot, which can help in understanding the behavior of the function.
  • If the derivative is positive, the function is increasing at that point.
  • If the derivative is negative, the function is decreasing.
  • If the derivative is zero, the function might have a peak or a low point, known as turning points.
Understanding derivatives is essential not just in mathematics but also in physics, engineering, and economics where they're used to model and predict various phenomena.
Power Rule
The Power Rule is one of the fundamental techniques for finding derivatives quickly. It's about formulating a simple rule to differentiate terms where the variable is raised to a power, given by the expression \(x^n\). According to the power rule,
If you have a function \(f(x) = x^n\), then its derivative is found using the formula:
\(f'(x) = nx^{n-1}\).
For example, if you need to find the derivative of \(-x^2\), you apply the Power Rule and get \(-2x\). The power \(n\) here is \(2\), and following the rule, you multiply by \(n\) and subtract one from the power.
Remembering the power rule simplifies the process of differentiation significantly, enabling you to quickly tackle polynomial functions that pop up across various problems and equations.
Exponential Functions
Exponential functions are functions where the variable is an exponent. In many cases, they involve the mathematical constant \(e\), approximately equal to 2.718. These functions have unique properties that make them significant in both pure and applied mathematics. They appear in numerous real-world applications like population growth, interest calculations, and complex chemical reactions.
One essential aspect of exponential functions is their rate of growth; they grow rapidly compared to polynomial functions. When it comes to differentiation, an important property of \(e^x\) is that its derivative is itself, \(e^x\).
For example, in the function \(y(e^u)\), the derivative with respect to \(u\) is still \(e^u\). This is due to the natural exponential function's unique characteristic that it doesn't change when differentiated, making calculations straightforward and often just involving multiplication by the derivative of the inner function, as seen in the Chain Rule application. This property slightly simplifies solving derivatives involving exponential functions.
  • Rapid growth and self-derivative make exponential functions standout.
  • They are widely used in continuous compound interest and vulnerability modeling in finance and other disciplines.

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Most popular questions from this chapter

An observer is \(20 \mathrm{m}\) above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is \(20 \mathrm{m}\) horizontally from the observer (see figure). The angle of elevation of the elevator is the angle that the observer's line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator rises at a rate of \(5 \mathrm{m} / \mathrm{s}\), what is the rate of change of the angle of elevation when the elevator is \(10 \mathrm{m}\) above the ground? When the elevator is \(40 \mathrm{m}\) above the ground?

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

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}(\ln x)^{x^{2}}$$

Assuming the first and second derivatives of \(f\) and \(g\) exist at \(x\), find a formula for \(\frac{d^{2}}{d x^{2}}(f(x) g(x))\)

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals). A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=c x^{2} ; x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants
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