Problem 29 - Single Variable Calculus

Question

Use logarithmic differentiation to find the derivative of the function. $$ y=(2 x+1)^{2}\left(3 x^{2}-4\right)^{3} $$

Step-by-Step Solution

Answer

The derivative of the function \(y(x) = (2x+1)^{2}(3x^{2}-4)^{3}\) with respect to x is: \(y'(x) = (2x+1)^{2}(3x^2-4)^{3}\left[\frac{4}{2x+1} + \frac{18x}{3x^2-4}\right]\).

Step 1: Take the natural logarithm of both sides

Take the natural logarithm of both sides of the equation: $$ \ln{y(x)} = \ln{[(2x+1)^{2}(3x^2-4)^3]} $$

Step 2: Apply logarithmic properties

Using the logarithmic properties, simplify the equation: $$ \ln{y(x)} = 2\ln{(2x+1)} + 3\ln{(3x^2-4)} $$

Step 3: Differentiate both sides implicitly with respect to x

Differentiate both sides of the equation with respect to x: $$ \frac{1}{y(x)}\frac{dy(x)}{dx} = 2\frac{d}{dx}[\ln{(2x+1)}] + 3\frac{d}{dx}[\ln{(3x^2-4)}] $$ Using the chain rule for differentiation, we get: $$ \frac{1}{y(x)}\frac{dy(x)}{dx} = 2\frac{2}{2x+1} + 3\frac{6x}{3x^2-4} $$

Step 4: Solve for y'(x)

Now, we need to solve for the first derivative, y'(x), by multiplying both sides of the equation by y(x): $$ \frac{dy(x)}{dx} = y(x)\left[2\frac{2}{2x+1} + 3\frac{6x}{3x^2-4}\right] $$ Substitute the original expression for y(x): $$ \frac{dy(x)}{dx} = (2x+1)^{2}(3x^2-4)^{3}\left[2\frac{2}{2x+1} + 3\frac{6x}{3x^2-4}\right] $$ Finally, simplify the expression: $$ \frac{dy(x)}{dx} = (2x+1)^{2}(3x^2-4)^{3}\left[\frac{4}{2x+1} + \frac{18x}{3x^2-4}\right] $$ Thus, the derivative of the function \(y(x)\) with respect to x is: $$ y'(x) = (2x+1)^{2}(3x^2-4)^{3}\left[\frac{4}{2x+1} + \frac{18x}{3x^2-4}\right] $$

Key Concepts

Chain Rule

The chain rule is a fundamental theorem that's pivotal for differentiating composite functions. In essence, it helps us calculate the derivative of functions nested inside one another. When you have a function inside another function, like \( f(g(x)) \), this rule is your go-to. To apply it, first find the derivative of the outer function evaluated at the inner function. Then, multiply this by the derivative of the inner function.
Memorize this formula for the chain rule:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)

For our situation, it's used with natural logarithms to differentiate each function inside the logs effectively. It's vital for solving problems involving multiple layers of functions, just like peeling an onion one layer at a time.

Implicit Differentiation

Implicit differentiation is a useful technique when dealing with functions that aren't isolated as \( y = f(x) \). Instead, you handle \( y \) and \( x \) being mixed together, such as in the relationship \( F(x, y) = 0 \). Instead of rearranging all the terms to get \( y \) alone, you differentiate with respect to \( x \) directly.
Here’s how to do it:

• Differentiating both sides while treating \( y \) as a function of \( x \).
• Remember to use the chain rule when you encounter \( y \).
• Solve the resulting equation for \( \frac{dy}{dx} \).

In our original exercise, after taking the natural log of both sides, we differentiated implicitly. This let us stay clear of rearranging terms excessively and directly find the derivative.

Derivative Calculation

Calculating derivatives can sometimes feel like solving a puzzle, especially with complex functions. But with the right rules, it can be straightforward. Start with:

• Applying the rules of differentiation, like the power rule and product rule.
• Using logarithmic differentiation, particularly effective for products, quotients, and powers, simplifies our task.

For instance, in the problem involving \((2x+1)^{2}(3x^2-4)^3\), directly computing the derivative can be tedious. Hence, logarithmic differentiation comes to aid by:- Allowing you to simplify expressions using the properties of logs.- Breaking down the original function into more manageable parts.By doing this, we differentiate simpler terms instead of a daunting, expanded product.

Natural Logarithms

Natural logarithms (denoted \( \ln \)) play a crucial role in calculus, especially in differentiation. They simplify the process of taking derivatives of products and powers through their properties. The function \( \ln(x) \) has an elegant property where the derivative is \( \frac{1}{x} \).
Here's why they're handy:

• Convert difficult multiplications into manageable additions.
• Transform powers into coefficient multipliers.

In logarithmic differentiation, natural logarithms let us handle each piece of a complex function independently before piecing them back together. This technique is especially useful for simplifying derivatives when you face functions like \((a^{b})\), allowing you to transform it into \( b \cdot \ln(a) \) and differentiate it more easily.