Problem 17 - Calculus Single Variable

Question

Calculate the derivative of the given expression with respect to \(x\). $$ \exp \left(-1 / x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative is \( \exp\left(-\frac{1}{x^2}\right) \cdot \frac{2}{x^3} \).

Step 1: Identify Outer and Inner Functions

The given expression is composed of two functions. The outer function is the exponential function, \( ext{exp}(u) \), and the inner function is \( u = -1/x^2 \). We'll apply the chain rule to find the derivative.

Step 2: Differentiate the Outer Function

The derivative of the outer function, \( ext{exp}(u) \), with respect to \( u \) is \( ext{exp}(u) \). Therefore, the derivative part of the outer function is: \[ \frac{d}{du} \exp(u) = \exp(u). \]

Step 3: Differentiate the Inner Function

Differentiate the inner function, \( u = -1/x^2 \). Rewriting this as \( u = -(x^{-2}) \), we apply the power rule: \[ \frac{du}{dx} = -(-2)x^{-3} = \frac{2}{x^3}. \]

Step 4: Apply the Chain Rule

Using the chain rule, the derivative of \( \exp(-1 / x^2) \) with respect to \( x \) is the product of the derivative of the outer function and the derivative of the inner function: \[ \frac{d}{dx} \exp(-1/x^2) = \exp(-1/x^2) \cdot \frac{2}{x^3}. \]

Step 5: Write the Final Answer

Combine the terms to express the final derivative: \[ \exp\left(-\frac{1}{x^2}\right) \cdot \frac{2}{x^3}. \] This is the derivative of \( \exp(-1/x^2) \) with respect to \( x \).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of composite functions. If you have a function like \( f(g(x)) \), where \( g(x) \) is nested inside another function \( f \), the chain rule is your go-to tool. It helps us find the derivative of the entire composite function rather than doing it separately, which would be impractical or impossible.To apply the chain rule, we differentiate the outer function concerning the inner function and multiply it by the derivative of the inner function. This process is expressed mathematically as:

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

In our original exercise, the outer function is the exponential and the inner function is \( -1/x^2 \). Following the chain rule:

• Differentiate the outer function: the derivative of \( \exp(u) \) with respect to \( u \) is \( \exp(u) \).
• Differentiate the inner function \( u = -1/x^2 \), which gives us \( 2/x^3 \) using the power rule.
• Multiply these two results to get the final derivative.

This process ensures we correctly account for how each part of the composite function changes as \( x \) changes.

Power Rule

The power rule is another essential rule in differentiation, mainly used when dealing with power functions of the form \( x^n \). It's one of the simplest and most useful rules for calculus students to master.To differentiate a power function \( x^n \), where \( n \) is any real number, you simply move the power to the front as a coefficient and subtract one from the power. Mathematically, this is:

• The derivative of \( x^n \) is \( nx^{n-1} \).

Using this rule, even negative and fractional powers can be differentiated easily. In our case, we have \( -1/x^2 \) which can be rewritten using negative exponents as \( -(x^{-2}) \).

• Applying the power rule, the derivative becomes \( 2x^{-3} \), or more familiarly, \( \frac{2}{x^3} \).

Mastering the power rule allows you to tackle a wide range of functions by simplifying the process of differentiation.

Exponential Function

Exponential functions are a class of functions where a constant base is raised to a variable exponent, often written as \( a^x \) or more commonly \( \exp(x) \), where \( \exp(x) = e^x \) and \( e \) is Euler’s number, approximately equal to 2.718.The beauty of exponential functions in calculus is their unique property: the derivative of \( e^x \) is itself, \( e^x \). This simplicity and elegance carry over to composite exponential functions via the chain rule.When dealing with functions like \( \exp(-1/x^2) \), the derivative with respect to that specific \( u \) (or inner function) stays the same. So:

• The derivative of \( \exp(u) \) where \( u \) is any differentiable function, is \( \exp(u) \), multiplied by the derivative of \( u \).

This makes exponential functions incredibly powerful and convenient to work with since they maintain their form upon differentiation. Thus, understanding this property is essential to efficiently solve complex problems involving exponential growth or decay.