Problem 16 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=x\left[\log _{2}\left(x^{2}-2 x\right)\right]^{3}$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = [\log_2(x^2 - 2x)]^3 + \frac{6x(x-1)[\log_2(x^2 - 2x)]^2}{(x^2 - 2x)\ln(2)}\)

Step 1: Recognize the Rule Needed

The function given is of the form \( y = x [f(x)]^3 \), where \( f(x) = \log_2(x^2 - 2x) \). We need to apply both the product rule and the chain rule to differentiate it.

Step 2: Apply the Product Rule

To differentiate a function of the form \( u(x) v(x) \), we use the product rule: \( \frac{d}{dx}[u(x) v(x)] = u'(x)v(x) + u(x)v'(x) \). Here, let \( u(x) = x \) and \( v(x) = [f(x)]^3 \). First, find \( u'(x) = 1 \).

Step 3: Differentiate \( v(x) \) Using the Chain Rule

\( v(x) = [f(x)]^3 \), hence \( v'(x) = 3[f(x)]^2 \cdot f'(x) \). The chain rule gives the derivative of a composition of functions. Now, find \( f'(x) \).

Step 4: Differentiate \( f(x) \) Using the Chain Rule and Logarithmic Derivative

\( f(x) = \log_2(x^2 - 2x) \) can be rewritten using the change of base formula as \( f(x) = \frac{\ln(x^2 - 2x)}{\ln(2)} \). The derivative is \( f'(x) = \frac{1}{\ln(2)} \cdot \frac{1}{x^2 - 2x} \cdot (2x - 2) \). Simplify to \( f'(x) = \frac{2(x-1)}{(x^2 - 2x)\ln(2)} \).

Step 5: Combine All Components

Combine the components from steps 2, 3, and 4 into the product rule formula. Thus, \(\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 1 \cdot [\log_2(x^2 - 2x)]^3 + x \cdot 3[\log_2(x^2 - 2x)]^2 \cdot \frac{2(x-1)}{(x^2 - 2x)\ln(2)}\).

Step 6: Simplify the Expression

Simplify the expression:\[\frac{dy}{dx} = [\log_2(x^2 - 2x)]^3 + \frac{6x(x-1)[\log_2(x^2 - 2x)]^2}{(x^2 - 2x)\ln(2)}\]

Key Concepts

Product Rule

The product rule is essential in calculus for differentiating expressions where two functions are multiplied together. If you have a function given by the product of two functions, say \( u(x) \) and \( v(x) \), the derivative is not just the product of their individual derivatives. Instead, the product rule states that the derivative of their product is:

• \( \frac{d}{dx}[u(x) v(x)] = u'(x)v(x) + u(x)v'(x) \)

To apply this, you must identify which part of your expression is \( u(x) \) and which part is \( v(x) \). For example, in the function \( y = x[f(x)]^3 \), we chose \( u(x) = x \) and \( v(x) = [f(x)]^3 \). Following the product rule, you differentiate \( u(x) \) to get \( u'(x) = 1 \), and then find \( v'(x) \) using other rules, which we will delve into in the next sections.

Chain Rule

The chain rule comes into play when you need to differentiate a composite function. A composite function is where one function is nested inside another, like \( v(x) = [f(x)]^3 \). The chain rule helps you differentiate such a function by following these steps:

• First, take the derivative of the outer function while keeping the inner function unchanged.
• Then, multiply that derivative by the derivative of the inner function.

For \( v(x) = [f(x)]^3 \), you would differentiate the outer function, \( x^3 \), yielding \( 3[f(x)]^2 \), and then multiply it by the derivative of \( f(x) \). Hence, \( v'(x) = 3[f(x)]^2 \cdot f'(x) \). This rule is powerful and widely applicable wherever you encounter layers of functions.

Logarithmic Differentiation

Logarithmic differentiation is a technique used when dealing with complex functions, especially when they involve logarithms. In our original exercise, we need to differentiate a function involving a log base 2: \( f(x) = \log_2(x^2 - 2x) \). However, it's simpler to use natural logarithms (\( \ln \)) in differentiation. Using the change of base formula, \( \log_2(x^2 - 2x) \) can be rewritten as \( \frac{\ln(x^2 - 2x)}{\ln(2)} \). To differentiate this, first take the derivative of \( \ln(x^2 - 2x) \) which gives:\( \frac{1}{x^2 - 2x} \cdot (2x - 2) \). Then divide by \( \ln(2) \) to adjust for the original base.of. The derivative of \( f(x) \) becomes \( f'(x) = \frac{2(x-1)}{(x^2 - 2x)\ln(2)} \). This method allows for simpler calculations even when faced with complex logarithmic expressions.