Problem 3 - Calculus

Question

Find the derivative \(d y / d x\) in \(1-10\). $$ y=(\ln x)^{-1} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = - \frac{1}{x(\ln x)^2} \).

Step 1: Rewrite the expression in a more convenient form

The given expression is \( y = (\ln x)^{-1} \). We can rewrite this as \( y = (\ln x)^{-1} = \ln x^{-1} \).

Step 2: Apply the chain rule

To find the derivative \( \frac{dy}{dx} \) of \( y = (\ln x)^{-1} \), use the chain rule for derivatives. The derivative of \( u^{-1} \) with respect to \( u \) is \( -u^{-2} \), and the derivative of \( \ln x \) with respect to \( x \) is \( \frac{1}{x} \). Therefore, \( \frac{dy}{dx} = - (\ln x)^{-2} \cdot \frac{1}{x} \).

Step 3: Simplify the expression

Now, simplify the expression for the derivative: \( \frac{dy}{dx} = - \frac{1}{x(\ln x)^2} \).

Key Concepts

chain rule

When you have a complex function that is a combination of two or more basic functions, the chain rule becomes a crucial differentiation technique. The chain rule helps find the derivative of such composite functions by breaking them down into simpler parts. Consider the function \( y = (\ln x)^{-1} \). This involves composing the function \( (\ln x) \) with the power \( -1 \).

• First, identify the outer function and the inner function. Here, the outer function is \( u^{-1} \) and the inner function is \( u = \ln x \).
• Differentiate the outer function with respect to the inner function: the derivative of \( u^{-1} \) is \( -u^{-2} \).
• Next, differentiate the inner function, \( \ln x \), with respect to \( x \), which gives \( \frac{1}{x} \).
• Finally, combine these derivatives by multiplying them: \( \frac{dy}{dx} = - (\ln x)^{-2} \cdot \frac{1}{x} \).

With this process, the chain rule elegantly handles the derivative of a function within another function.

logarithmic functions

Logarithmic functions, which often involve expressions like \( \ln x \), are essential in calculus, especially when handling exponential growth, decay, and solving complex differential equations. The natural logarithm, denoted as \( \ln \), is the inverse function of the exponential function \( e^x \). It has several important properties useful in differentiation:

• Its domain is \( x > 0 \), meaning you can only plug positive numbers into \( \ln x \).
• The derivative of \( \ln x \) with respect to \( x \) is \( \frac{1}{x} \).
• Logarithms have useful identities like \( \ln(a^b) = b\ln a \), which can assist in simplifying complex expressions.

Understanding these properties helps when calculating derivatives involving logarithms, as it provides straightforward rules to follow. This makes analyzing and differentiating logarithmic functions more manageable in calculus problems.

differentiation techniques

Differentiation techniques are crucial tools in calculus for finding the rate at which functions change. There are several key methods and formulas for differentiation, each suited to handle different types of functions. Here’s an overview:

• Power Rule: For any function \( f(x) = x^n \), its derivative is \( f'(x) = nx^{n-1} \), where \( n \) can be any real number. It simplifies derivatives of polynomial functions.
• Product Rule: Used when differentiating products of two functions, it states that \( (uv)' = u'v + uv' \).
• Quotient Rule: Applied to the quotient of two functions, given by \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).
• Chain Rule: This crucial rule deals with composite functions like \( f(g(x)) \), allowing us to differentiate them by intertwining their respective derivatives effectively.

Each technique can be applied to tackle different challenges in finding derivatives. Mastery of these techniques not only enables smoother computation but also deepens understanding of how functions behave and change.