Question

T/F: \(\frac{d}{d x}\left(\ln \left(x^{2}\right)\right)=\frac{1}{x^{2}}\).

Short answer

False, the derivative is \(\frac{2}{x}\), not \(\frac{1}{x^2}\).

Step-by-step solution

Identify the Function and Apply the Chain Rule

The function given is \(y = \ln(x^2)\). To differentiate this, we need to apply the chain rule. The chain rule states that if a function \(y = f(g(x))\) is given, then its derivative is \(f'(g(x)) \cdot g'(x)\).

Find the Outer Function's Derivative

In this case, the outer function is \(f(u) = \ln(u)\) where \(u = x^2\). The derivative of \(\ln(u)\) with respect to \(u\) is \(1/u\). Thus, \(f'(u) = \frac{1}{u}\).

Find the Inner Function's Derivative

The inner function is \(g(x) = x^2\). The derivative of \(x^2\) with respect to \(x\) is \(2x\). So, \(g'(x) = 2x\).

Apply the Chain Rule to Differentiate

Using the chain rule, the derivative \(\frac{d}{dx} (\ln(x^2))\) is calculated as:\[ \frac{d}{dx} (\ln(x^2)) = f'(g(x)) \cdot g'(x) = \frac{1}{x^2} \cdot 2x = \frac{2x}{x^2} = \frac{2}{x}. \]

Compare the Result to the Given Expression

The derivative found is \(\frac{2}{x}\), whereas the given expression was \(\frac{1}{x^2}\). These two expressions are clearly different.

Key concepts

Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of composite functions. It allows you to differentiate functions inside other functions with ease. Think of a composite function as a function inside another function. In this context, the problem involves \(y = \ln(x^2)\) where \(x^2\) is inside \(\ln\).To use the chain rule, follow these two main steps:

• Find the derivative of the outer function: Consider the outer function as if it only has a variable \(u\), for example, \(f(u) = \ln(u)\). Differentiate \(f(u)\) with respect to \(u\) to get \(f'(u) = \frac{1}{u}\).
• Find the derivative of the inner function: The inner function here is \(u = x^2\), and its derivative is \(g'(x) = 2x\).

Now, multiply the derivatives of these functions together. This gives:\[ f'(g(x)) \cdot g'(x) = \frac{1}{x^2} \cdot 2x = \frac{2}{x}. \]Notice how each step uses both the inner and outer function derivatives. This step-by-step approach reveals the power and simplicity of the chain rule.

Logarithmic Differentiation

Logarithmic differentiation is particularly useful when dealing with logarithmic functions and powers. It simplifies complex products and quotients, and can even convert multiplicative processes into additive ones, making differentiation easier.In the context of our exercise, \(y = \ln(x^2)\) is the function. By understanding the properties of logarithms, you can transform difficult problems into simpler ones. Here converting the exponent to multiplication is achieved by the logarithmic identity:

• \( \ln(x^n) = n \ln(x) \)

Applying this property, rewrite \(y = \ln(x^2) \) to \(2 \ln(x)\). Differentiating \(2 \ln(x)\) directly gives:\[ \frac{d}{dx}(2 \ln(x)) = 2 \cdot \frac{1}{x} = \frac{2}{x}. \]This alternative approach also leads to the same derivative as using the chain rule, showing the versatility of logarithmic differentiation. It can sometimes simplify differentiation, especially when a function is complex in its initial form.

Function Derivative

Understanding the derivative of a function is fundamental in calculus. A derivative represents how a function changes as its input changes. Essentially, it gives the rate of change or the slope of the function at any given point.In the exercise, the function we differentiate is \(y = \ln(x^2)\). It is layered with inner and outer functions, making it an ideal candidate for differentiation techniques like the chain rule or logarithmic differentiation. The derivative of the outer function \(\ln(u)\) is \(\frac{1}{u}\), while the derivative of the inner function \(x^2\) is \(2x\).Putting these steps together, applying the chain rule yields the final derivative:\[ \frac{d}{dx} (\ln(x^2)) = \frac{2}{x}. \]Knowing how to derive these expressions reveals the underlying patterns in calculus and aids in solving diverse problems involving derivatives. Mastery of these concepts will enable you to tackle various functions with confidence.