Problem 31 - Calculus - AP Edition

Question

Find \(d y / d x\) using the method of logarithmic differentiation. $$y=\left(x^{3}-2 x\right)^{\ln x}$$

Step-by-Step Solution

Answer

Use logarithmic differentiation to find \( \frac{dy}{dx} = (x^3 - 2x)^{\ln x} \left( \frac{\ln(x^3 - 2x)}{x} + \frac{\ln x \cdot (3x^2 - 2)}{x^3 - 2x} \right) \).

Step 1: Take the Natural Logarithm of Both Sides

Start by taking the natural logarithm of both sides of the equation. This gives you: \[ \ln y = \ln\left( (x^3 - 2x)^{\ln x} \right) \]Using the logarithm power rule, \( \ln(a^b) = b \ln a \), you can simplify this to: \[ \ln y = (\ln x) \cdot \ln(x^3 - 2x) \]

Step 2: Differentiate Both Sides Implicitly with Respect to x

Differentiate both sides with respect to \(x\). For the left side, the derivative of \( \ln y \) with respect to \( x \) is \( \frac{1}{y} \frac{dy}{dx} \). For the right side, use the product rule \( (uv)' = u'v + uv' \):\[ \frac{d}{dx}(\ln x \cdot \ln(x^3 - 2x)) = (\frac{1}{x})(\ln(x^3 - 2x)) + (\ln x) \cdot \frac{d}{dx}(\ln(x^3 - 2x)) \]The derivative of \( \ln(x^3 - 2x) \) is \( \frac{1}{x^3 - 2x} \cdot (3x^2 - 2) \) using the chain rule and power rule.

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

Combine the results from the differentiation:\[ \frac{1}{y} \frac{dy}{dx} = \frac{\ln(x^3 - 2x)}{x} + \frac{\ln x \cdot (3x^2 - 2)}{x^3 - 2x} \]Multiply through by \( y \) to solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = y \left( \frac{\ln(x^3 - 2x)}{x} + \frac{\ln x \cdot (3x^2 - 2)}{x^3 - 2x} \right) \]

Step 4: Substitute Back for y

Remembering that \( y = (x^3 - 2x)^{\ln x} \), substitute \( y \) back into the equation:\[ \frac{dy}{dx} = (x^3 - 2x)^{\ln x} \left( \frac{\ln(x^3 - 2x)}{x} + \frac{\ln x \cdot (3x^2 - 2)}{x^3 - 2x} \right) \]

Key Concepts

Implicit Differentiation

The journey of using logarithmic differentiation starts with implicit differentiation. Implicit differentiation is a technique used when a function is not given in the explicit format of \( y = f(x) \). Instead, the function might be intertwined, leaving \( y \) embedded among other terms or within complicated expressions. This prevents us from easily expressing \( y \) solely with \( x \).

In the context of the addressed exercise, implicit differentiation is applied after the natural logarithm is taken. The given equation \( y = (x^3 - 2x)^{\ln x} \) undergoes transformation to its logarithmic form. This renders:\[\ln y = (\ln x) \cdot \ln(x^3 - 2x)\]

Given this transformation, implicit differentiation clears up the path. It allows for differentiating both sides with respect to \( x \). The left side of the equation, which now stands as \( \ln y \), can be differentiated to yield: \( \frac{1}{y} \frac{dy}{dx} \). Meanwhile, the right side requires further steps provided by the product and chain rule, which we will delve into next.

Chain Rule

The chain rule is a fundamental aspect of calculus that simplifies deriving complex functions. It is essential when a function is nested within another function. The crux of this rule is to "differentiate the outer function and multiply by the derivative of the inner function."

Returning to our exercise, the chain rule becomes necessary for differentiating \( \ln(x^3 - 2x) \). This component is a mixture of several functions: the logarithm is the outer function, while \( x^3 - 2x \) is the inner function. Applying the chain rule, we differentiate the logarithm and multiply by the derivative of the polynomial \( x^3 - 2x \). Here's the breakdown:

• The derivative of \( \ln u \) is \( \frac{1}{u} \).
• In our case, \( u = x^3 - 2x \). Thus, substitute to obtain \( \frac{1}{x^3 - 2x} \).
• The derivative of \( x^3 - 2x \) is \( 3x^2 - 2 \).
• Combine these results to find \( \frac{d}{dx}(\ln(x^3 - 2x)) = \frac{1}{x^3 - 2x} \cdot (3x^2 - 2) \).

By mastering the chain rule, differentiation of such composed functions becomes systematic and straightforward.

Product Rule

When dealing with multiplication of functions, the product rule is an indispensable tool. The product rule dictates that the derivative of a product \( uv \) is \( u'v + uv' \). This rule applies when you need to find the derivative of one function multiplied by another.

In this exercise, after taking logs and setting up implicit differentiation, the right side of the equation becomes \( \ln x \cdot \ln(x^3 - 2x) \). To differentiate this, the product rule is necessary since it's the product of two distinct functions: \( u = \ln x \) and \( v = \ln(x^3 - 2x) \).

Follow the steps below to implement the product rule for our scenario:

• Differentiate \( u = \ln x \) to obtain \( u' = \frac{1}{x} \).
• Using the chain rule, we've already found that \( v' = \frac{3x^2 - 2}{x^3 - 2x} \).
• Apply the product rule: \( u'v + uv' \) which translates to \( \frac{1}{x}\ln(x^3 - 2x) + \ln x \cdot \frac{3x^2 - 2}{x^3 - 2x} \).

Employing the product rule in this manner ensures that differentiating combinations of products within complicated expressions is tackled effectively.