Problem 18 - Calculus

Question

Find the derivatives of the functions $$ x^{-1 / x} \text { (which is } e-\text { ) } $$

Step-by-Step Solution

Answer

The derivative of \(x^{-1/x}\) is \(x^{-1/x}\frac{\ln(x) + 1}{x^2}\).

Step 1: Rewrite the Function

First, rewrite the function for convenience. Let \( y = x^{-1/x} \). Then take the natural logarithm of both sides to get \( \ln(y) = \ln(x^{-1/x}) \). This simplifies to \( \ln(y) = -\frac{1}{x} \ln(x) \).

Step 2: Differentiate Implicitly

Differentiate both sides with respect to \( x \). Using the chain rule on the left, \( \frac{d(\ln(y))}{dx} = \frac{1}{y} \frac{dy}{dx} \). On the right, use the product rule: \( \frac{d}{dx} \left(-\frac{1}{x} \ln(x)\right) = -\left(\frac{d}{dx}(\frac{1}{x})\cdot\ln(x) + \frac{1}{x}\cdot\frac{d}{dx}(\ln(x))\right) \). This simplifies to \( \frac{1}{y} \frac{dy}{dx} = \frac{\ln(x) + 1}{x^2} \).

Step 3: Solve for the Derivative \( \frac{dy}{dx} \)

Now solve for \( \frac{dy}{dx} \) by multiplying both sides by \( y \), giving \( \frac{dy}{dx} = y \left( \frac{\ln(x) + 1}{x^2} \right) \). Substitute back \( y = x^{-1/x} \) to get \( \frac{dy}{dx} = x^{-1/x} \left( \frac{\ln(x) + 1}{x^2} \right) \).

Step 4: Simplify the Expression

The derivative \( \frac{dy}{dx} = x^{-1/x} \frac{\ln(x) + 1}{x^2} \) can be simplified and used as needed. However, further simplification would involve messy algebra without additional functional requirements, so it is often left as is.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations that define a relationship between variables, and it's more convenient to handle the differentiation without explicitly solving for one variable in terms of the other.
In the exercise, after taking the natural logarithm of the function, we obtained an equation involving both \( y \) and \( x \): \( \ln(y) = -\frac{1}{x} \ln(x) \). Since it's challenging to express \( y \) explicitly in simple terms for differentiation, we instead differentiate both sides with respect to \( x \) as they stand.

• On the left side, \( \frac{d(\ln(y))}{dx} \) involves the derivative of a logarithm of \( y \), which by the chain rule becomes \( \frac{1}{y} \frac{dy}{dx} \).
• On the right side, differentiation will require attention to both the numerator and the natural log due to \( \frac{1}{x} \ln(x) \).

Implicit differentiation allows us to solve for \( \frac{dy}{dx} \) without prior isolation of \( y \).

Natural Logarithm

The natural logarithm, also represented as \( \ln(x) \), is a fundamental mathematical function with special properties that make it highly useful in calculus and algebra.
For the exercise in question, we start with rewriting the original function \( y = x^{-1/x} \) using the natural logarithm: \( \ln(y) = \ln(x^{-1/x}) \). This simplifies to \( \ln(y) = -\frac{1}{x} \ln(x) \), taking advantage of the logarithmic identity: \( \ln(a^b) = b\ln(a) \).

• The natural logarithm transforms products, powers, and quotients into simpler sums, differences, and products, respectively, making differentiation easier.
• It's crucial to understand that \( \ln(x) \) is undefined for \( x \leq 0 \), which means our calculations assume \( x \) is positive.

These properties make the natural logarithm a powerful tool in manipulating and simplifying complex differentiation problems.

Chain Rule

When you differentiate a function composed of several nested functions, the chain rule becomes an indispensable tool. It provides a method for dealing with such composites systematically.
In the exercise, we encounter it primarily when differentiating \( \ln(y) \) with respect to \( x \), as \( y \) is itself a function of \( x \). Here, the chain rule gives us: \( \frac{d(\ln(y))}{dx} = \frac{1}{y} \frac{dy}{dx} \).

• The outer function is \( \ln(u) \) where \( u = y \), and its derivative is \( \frac{1}{u} \).
• The inner function is \( y \), which depends on \( x \). Here, the derivative we are looking for is \( \frac{dy}{dx} \).

By multiplying these derivatives, as the chain rule dictates, we ensure that the result takes into account all layers of functions involved.

Product Rule

Whenever a function is the product of two other functions, the product rule will guide us in effectively finding its derivative.
In the exercise’s context, after applying the natural logarithm \( -\frac{1}{x} \ln(x) \), we identify two functions within the product: \( -\frac{1}{x} \) and \( \ln(x) \).
The product rule can be stated mathematically as: for two functions \( u(x) \) and \( v(x) \), their derivative is given by \( u'(x)v(x) + u(x)v'(x) \).

• \( u(x) = -\frac{1}{x} \) and \( v(x) = \ln(x) \).
• Differentiating these using basic derivative rules: \( u'(x) = \frac{1}{x^2} \) and \( v'(x) = \frac{1}{x} \).
• Applying the product rule yields \( -\left( \frac{1}{x^2}\ln(x) + \frac{1}{x}\cdot\frac{1}{x} \right) \).

Thus, the product rule elegantly helps us manage differentiation where multiplying parts are involved.