Understanding the Derivative of ln x Squared

The derivative of the function \( \ln{(x^2)} \) can be derived using various methods such as the chain rule, first principles, implicit differentiation, and the product rule.

Using the Chain Rule

Let's define the function as:

\( f(x) = \ln{(x^2)} \)

We can rewrite it using the chain rule:

• \( u = x^2 \)
• \( f(u) = \ln{(u)} \)

Then, applying the chain rule:

\( \frac{dy}{dx} = f'(u) \cdot g'(x) \)

\( f'(u) = \frac{1}{u} \) and \( g'(x) = 2x \)

Thus:

\( \frac{dy}{dx} = \left( \frac{1}{x^2} \right) \cdot (2x) = \frac{2}{x} \)

Using Implicit Differentiation

Starting with:

\( y = \ln{(x^2)} \)

Convert to exponential form:

\( e^y = x^2 \)

Differentiate both sides with respect to \( x \):

\( e^y \cdot \frac{dy}{dx} = 2x \)

Isolate \( \frac{dy}{dx} \):

\( \frac{dy}{dx} = \frac{2x}{e^y} \)

Substitute back \( e^y = x^2 \):

\( \frac{dy}{dx} = \frac{2x}{x^2} = \frac{2}{x} \)

Using the Product Rule

Rewrite \( \ln{(x^2)} \) using logarithm properties:

\( \ln{(x^2)} = 2 \ln{(x)} \)

Differentiate using the product rule:

\( \frac{d}{dx} [2 \ln{(x)}] = 2 \cdot \frac{d}{dx} [\ln{(x)}] = 2 \cdot \frac{1}{x} = \frac{2}{x} \)

Graphical Interpretation

The graph of \( \ln{(x^2)} \) is a logarithmic curve that reflects the logarithm's nature, and its derivative \( \frac{2}{x} \) represents a hyperbola indicating the rate of change at each point on the original curve.

The derivative of ln(x^2) is a fundamental concept in calculus, crucial for understanding the rate of change in logarithmic functions. This section delves into the derivative of ln(x^2), illustrating its importance and applications in various scientific and engineering fields. We will explore different methods to derive this derivative, including the first principle, implicit differentiation, and the product rule.

The derivative of the natural logarithm of \(x^2\) can be derived using two methods: the chain rule and properties of logarithms.

1. Using the Chain Rule:

   The chain rule states that the derivative of a composite function \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\). Let \(y = \ln(x^2)\). We can set up our functions as follows:

   • Outer function: \(f(u) = \ln(u)\)
   • Inner function: \(u = x^2\)

   First, differentiate the outer function with respect to \(u\):

   \(f'(u) = \frac{1}{u}\)

   Then, differentiate the inner function with respect to \(x\):

   \(g'(x) = 2x\)

   Now, apply the chain rule:

   \(\frac{dy}{dx} = f'(u) \cdot g'(x) = \frac{1}{x^2} \cdot 2x = \frac{2x}{x^2} = \frac{2}{x}\)

2. Using Logarithm Properties:

   The logarithm property \(\ln(a^b) = b \cdot \ln(a)\) allows us to rewrite \(\ln(x^2)\) as:

   \(\ln(x^2) = 2 \cdot \ln(x)\)

   Then, differentiate with respect to \(x\):

   \(\frac{d}{dx} \left(2 \cdot \ln(x)\right) = 2 \cdot \frac{d}{dx} (\ln(x)) = 2 \cdot \frac{1}{x} = \frac{2}{x}\)

Therefore, the derivative of \(\ln(x^2)\) is \(\frac{2}{x}\).

The derivative of the natural logarithm function \(\ln(x^2)\) can be proven using different methods. Here we will explore the chain rule and implicit differentiation.

• Using the Chain Rule:
  1. Identify the outer function \( f(u) = \ln(u) \) and the inner function \( g(x) = x^2 \).
  2. Compute the derivative of the outer function: \( f'(u) = \frac{1}{u} \).
  3. Compute the derivative of the inner function: \( g'(x) = 2x \).
  4. Apply the chain rule: \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) = \frac{1}{x^2} \cdot 2x = \frac{2x}{x^2} = \frac{2}{x} \).
• Using Implicit Differentiation:
  1. Rewrite the function: \( y = \ln(x^2) \).
  2. Express in exponential form: \( e^y = x^2 \).
  3. Differentiate both sides with respect to \( x \): \( e^y \cdot \frac{dy}{dx} = 2x \).
  4. Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{2x}{e^y} \).
  5. Recall \( y = \ln(x^2) \), so \( e^y = x^2 \). Substituting gives: \( \frac{dy}{dx} = \frac{2x}{x^2} = \frac{2}{x} \).

Both methods confirm that the derivative of \( \ln(x^2) \) is \( \frac{2}{x} \).

To prove the derivative of \( \ln(x^2) \) using the first principles method, we start by applying the definition of the derivative:

\[
f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h}
\]

For \( f(x) = \ln(x^2) \), we have:

\[
f'(x) = \lim_{{h \to 0}} \frac{{\ln((x+h)^2) - \ln(x^2)}}{h}
\]

Using logarithm properties, we can simplify the expression:

\[
f'(x) = \lim_{{h \to 0}} \frac{{\ln((x+h)^2/x^2)}}{h}
\]

Simplifying further:

\[
f'(x) = \lim_{{h \to 0}} \frac{{\ln((x+h)/x)^2}}{h}
\]

\[
f'(x) = \lim_{{h \to 0}} \frac{{2\ln(1+h/x)}}{h}
\]

Let \( t = \frac{h}{x} \), thus \( h = tx \) and when \( h \to 0 \), \( t \to 0 \):

\[
f'(x) = \lim_{{t \to 0}} \frac{{2\ln(1+t)}}{tx}
\]

We can separate the constants:

\[
f'(x) = \frac{2}{x} \lim_{{t \to 0}} \frac{\ln(1+t)}{t}
\]

We know that \( \lim_{{t \to 0}} \frac{\ln(1+t)}{t} = 1 \):

\[
f'(x) = \frac{2}{x}
\]

Thus, the derivative of \( \ln(x^2) \) is:

\[
f'(x) = \frac{2}{x}
\]

The chain rule is an essential technique in calculus for finding the derivative of a composite function. In this section, we will prove the derivative of \( \ln(x^2) \) using the chain rule. Let's break it down step by step.

1. Identify the outer and inner functions: For \( \ln(x^2) \), the outer function is \( \ln(u) \) where \( u = x^2 \).
2. Differentiate the outer function with respect to the inner function: The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \).
3. Differentiate the inner function with respect to \( x \): The derivative of \( x^2 \) with respect to \( x \) is \( 2x \).
4. Apply the chain rule: Multiply the derivative of the outer function by the derivative of the inner function.
   • The derivative of \( \ln(u) \) with respect to \( x \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).
   • Substitute \( u = x^2 \) and \( \frac{du}{dx} = 2x \):
   • \( \frac{d}{dx} \ln(x^2) = \frac{1}{x^2} \cdot 2x = \frac{2x}{x^2} = \frac{2}{x} \).

Therefore, the derivative of \( \ln(x^2) \) using the chain rule is \( \frac{2}{x} \).

To prove the derivative of \( \ln(x^2) \) using implicit differentiation, follow these steps:

1. Start with the equation: \[ y = \ln(x^2) \]
2. Rewrite the equation in exponential form: \[ e^y = x^2 \]
3. Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(e^y) = \frac{d}{dx}(x^2) \]
4. Apply the chain rule on the left side and differentiate the right side: \[ e^y \cdot \frac{dy}{dx} = 2x \]
5. Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{2x}{e^y} \]
6. Substitute \( y = \ln(x^2) \) back into the equation: \[ \frac{dy}{dx} = \frac{2x}{e^{\ln(x^2)}} \]
7. Simplify using the property \( e^{\ln(a)} = a \): \[ \frac{dy}{dx} = \frac{2x}{x^2} \]
8. Simplify the fraction: \[ \frac{dy}{dx} = \frac{2}{x} \]

Thus, the derivative of \( \ln(x^2) \) is:
\[
\frac{d}{dx} \ln(x^2) = \frac{2}{x}
\]

The product rule states that the derivative of a product of two functions is given by:

\[
\frac{d}{dx}[f(x) \cdot g(x)] = f(x) \cdot \frac{d}{dx}g(x) + g(x) \cdot \frac{d}{dx}f(x)
\]

To apply the product rule to \( \ln(x^2) \), we can rewrite \( \ln(x^2) \) using the logarithm property that states \( \ln(x^2) = 2\ln(x) \). Therefore, we need to find the derivative of \( 2\ln(x) \).

Let \( u(x) = 2 \) and \( v(x) = \ln(x) \).

Now we apply the product rule:

1. First, differentiate \( u(x) \): \[ \frac{d}{dx}[2] = 0 \]
2. Next, differentiate \( v(x) \): \[ \frac{d}{dx}[\ln(x)] = \frac{1}{x} \]

Substitute these into the product rule formula:

\[
\frac{d}{dx}[2 \cdot \ln(x)] = 2 \cdot \frac{d}{dx}[\ln(x)] + \ln(x) \cdot \frac{d}{dx}[2]
\]

Since \( \frac{d}{dx}[2] = 0 \), this simplifies to:

\[
2 \cdot \frac{1}{x} + \ln(x) \cdot 0 = \frac{2}{x}
\]

Therefore, the derivative of \( \ln(x^2) \) using the product rule is:

\[
\frac{d}{dx}\ln(x^2) = \frac{2}{x}
\]

The derivative of \( \ln(x^2) \) is a fundamental concept in calculus with a wide range of applications across various fields. Understanding this derivative, which is \( \frac{2}{x} \), allows us to analyze and model complex phenomena effectively.

Physics and Engineering

• Aviation: The derivative helps in calculating the rate of climb or descent of an aircraft, which is crucial for navigation and flight safety.
• Electronics: In analyzing RC (resistor-capacitor) circuits, the derivative is used to determine the charging and discharging rates of capacitors, aiding in the design of efficient electronic systems.
• Acoustics: Sound pressure levels in decibels are calculated using logarithmic functions, and the derivative aids in the design and optimization of audio equipment and noise control measures.

Chemistry

In chemical kinetics, the rate of reaction and the change in concentration of reactants or products over time are analyzed using derivatives. This helps chemists predict reaction behavior and optimize conditions for desired outcomes.

Economics and Finance

• Economic Modeling: Derivatives help in predicting market trends, interest rates, and investment returns, guiding financial decisions and policy making.
• Optimization: Businesses use derivatives to optimize production and cost functions, improving efficiency and profitability.

Environmental Science

Modeling environmental data such as temperature variations and pollution levels often involves derivatives. These models are crucial for research on climate change and ecological impacts, helping scientists develop strategies to mitigate adverse effects.

These applications demonstrate the versatility and importance of understanding the derivative of \( \ln(x^2) \). It bridges theoretical mathematics with practical solutions to complex problems across various disciplines.

Graphical representation of the function \( \ln(x^2) \) and its derivative \( \frac{2}{x} \) can provide a clear visual understanding of their behaviors. Below, we will illustrate these graphs and discuss their characteristics.

Graph of \( \ln(x^2) \)

The function \( \ln(x^2) \) is defined for all real numbers except \( x = 0 \). The graph of \( \ln(x^2) \) is symmetric with respect to the y-axis because \( \ln(x^2) = \ln((-x)^2) \). As \( x \) approaches 0 from either direction, \( \ln(x^2) \) decreases without bound.

Here is the graph of \( \ln(x^2) \):

Graph of \( \frac{2}{x} \)

The derivative of \( \ln(x^2) \), given by \( \frac{2}{x} \), also has a distinct graphical representation. This function is defined for all \( x \neq 0 \). The graph of \( \frac{2}{x} \) has a vertical asymptote at \( x = 0 \) and horizontal asymptotes as \( x \) approaches \( \pm \infty \).

Here is the graph of \( \frac{2}{x} \):

Comparing the Graphs

By comparing the graphs of \( \ln(x^2) \) and \( \frac{2}{x} \), we can observe the following:

• Both functions are undefined at \( x = 0 \).
• \( \ln(x^2) \) has a local minimum at \( x = 1 \) and \( x = -1 \), where it equals 0.
• The derivative \( \frac{2}{x} \) changes sign at \( x = 0 \), indicating a critical point for \( \ln(x^2) \).
• The symmetry of \( \ln(x^2) \) about the y-axis reflects in the behavior of its derivative, which is an odd function.

These graphical insights help in understanding the relationship between a function and its derivative, showing how changes in the function are reflected in the slope of the tangent line at any given point.