Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ D_{x}\left(e^{1 / x^{2}}+1 / e^{x^{2}}\right) $$

Step-by-Step Solution

Answer

The derivative is \(-\frac{2}{x^3}e^{1/x^2} - 2xe^{-x^2}\).

Step 1: Differentiate the First Term

We first differentiate the function with respect to \( x \). Let's begin with the term \( e^{1/x^2} \). To differentiate \( e^{1/x^2} \), let \( u = 1/x^2 \), so \( du/dx = -2/x^3 \). The derivative of \( e^u \) with respect to \( u \) is \( e^u \), so by the chain rule, the derivative of \( e^{1/x^2} \) is \( e^{1/x^2} \cdot (-2/x^3) \).

Step 2: Differentiate the Second Term

Next, differentiate the second term \( 1/e^{x^2} \), which can be rewritten as \( e^{-x^2} \). Let \( v = x^2 \), so \( dv/dx = 2x \). The derivative of \( e^{-v} \) is \( -e^{-v} \cdot dv/dx \). Therefore, the derivative of \( e^{-x^2} \) is \( -e^{-x^2} \cdot 2x \).

Step 3: Combine the Derivatives

Combine the derivatives of each term: from Step 1 we have \(-\frac{2}{x^3}e^{1/x^2}\) and from Step 2 we have \( -2xe^{-x^2} \). Hence, the derivative \( D_x(e^{1/x^2} + 1/e^{x^2}) \) is \(-\frac{2}{x^3}e^{1/x^2} - 2xe^{-x^2} \).

Key Concepts

Derivative

Derivatives are fundamental in calculus. They measure how a function changes as its input changes. In simpler terms, the derivative gives us the rate of change of a function at a particular point. It can tell us how steep a graph is or how fast something is increasing or decreasing.

When we talk about derivatives, we're often using notations like \( D_x \) or the prime symbol (\( f'(x) \)) to denote differentiation. So, if we say "find \( D_x(f) \)," we are looking for how the function \( f(x) \) changes with respect to \( x \).

• The notation \( D_x \) is specifically referring to differentiation with respect to \( x \).
• It helps in understanding the behavior and the rate at which a particular function or equation changes.

The derivative can reveal important information about a function, such as identifying maximum and minimum values, understanding concavity, and solving practical problems in physics, engineering, and other sciences.

Chain Rule

The chain rule is one of the most powerful techniques in calculus for finding derivatives. It is especially useful when dealing with composite functions, where one function is inside another. This rule essentially says that to differentiate a function that's made up of a composition \( g(f(x)) \), you multiply the derivative of the outer function \( g \) by the derivative of the inner function \( f \).

In symbolic terms, if \( h(x) = g(f(x)) \), then \( h'(x) = g'(f(x)) \cdot f'(x) \). It might sound complicated, but it breaks down a complex function differentiation into simpler parts you can solve step-by-step.

• In our problem, the term \( e^{1/x^2} \) required the chain rule. Here, we differentiated the outer function, \( e^u \), where \( u = \frac{1}{x^2} \).
• The derivative of \( e^u \) is \( e^u \cdot \frac{du}{dx} \), with \( \frac{du}{dx} = -\frac{2}{x^3} \).

Using the chain rule, you can manage complex functions more efficiently by breaking them down into simpler, differentiable components.

Exponential Function

Exponential functions are functions of the form \( e^x \), where \( e \) is a special mathematical constant approximately equal to 2.71828. These functions are significant in many fields because they describe exponential growth or decay, common in natural and financial contexts.

One of the main reasons exponential functions are interesting in calculus is due to their unique property: their derivative is proportional to the function itself. In simpler words, if \( y = e^x \), then \( \frac{dy}{dx} = e^x \).

• This property is visible in our problem where differentiating \( e^{1/x^2} \) begins with the same exponential term in its structure.
• The same property holds when the exponent is replaced by any function of \( x \), such as \( e^{-x^2} \).

Understanding how to work with exponential functions and their derivatives is crucial because they naturally model a variety of processes, from radioactive decay to population growth.