Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Differentiate the following. $$ y=\left(3 x^{2}+2\right)^{x} $$

Step-by-Step Solution

Answer

The derivative of \( y=\left(3 x^{2}+2\right)^{x} \) is \( y' = \left(3x^{2}+2\right)^{x} \left(\ln\left(3x^{2}+2\right) + x \left(\frac{1}{3x^{2}+2}\right) \left(6x\right)\right) \)

Step 1: Take natural log on both sides

Let's write the given function in logarithmic form to make it easier to differentiate.\( \ln(y) = x \ln\left(3x^{2}+2\right) \)

Step 2: Differentiate both sides with respect to x

Apply the product rule on the right side. The product rule states that the derivative of two functions \(f(x)\) and \(g(x)\) is \(f'(x)g(x) + f(x)g'(x)\).Differentiating, we get:\( y'\left(\frac{1}{y}\right) = \ln\left(3x^{2}+2\right) + x \left(\frac{1}{3x^{2}+2}\right) \left(6x\right) \)

Step 3: Solve for y'

Multiplying both sides by y gives us the derivative.\( y'= y \left(\ln\left(3x^{2}+2\right) + x \left(\frac{1}{3x^{2}+2}\right) \left(6x\right)\right) \)

Step 4: Substitute y in the equation

Substitute \( y= \left(3x^{2}+2\right)^{x} \) into the equation to get the final derivative.\( y' = \left(3x^{2}+2\right)^{x} \left(\ln\left(3x^{2}+2\right) + x \left(\frac{1}{3x^{2}+2}\right) \left(6x\right)\right) \)

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \(\ln\), is a mathematical operation that helps simplify complex multiplication and powers in calculus. When differentiating functions like \( y = (3x^2 + 2)^x \), logarithms can be incredibly useful. By taking the natural logarithm of both sides of an equation, you can transform power functions into a more manageable form. This transformation makes it easier to differentiate the function using calculus rules.

The natural logarithm has properties that make differentiation more straightforward:

• \( \ln(a^b) = b \cdot \ln(a) \), which allows breaking powers into simpler logs
• \( \ln(e) = 1 \) for the natural base \( e \)
• Derivative of \( \ln(x) \) is \( \frac{1}{x} \)

In our example, the initial step involved rewriting \( y \) using the logarithmic identity: \( \ln(y) = x \ln(3x^2 + 2) \). This step simplifies differentiation processes that would otherwise be more complex.

Product Rule

The product rule is a key tool in differentiation used when dealing with the derivative of products of two functions. It states that if you have two functions, \( f(x) \) and \( g(x) \), their derivative, \( (f \cdot g)'(x) \), is given by:

\[ f'(x)g(x) + f(x)g'(x) \]

This rule is pivotal because many functions are products and need decomposition through this rule to differentiate accurately. For our function \( \ln(y) = x \ln(3x^2 + 2) \), you can see the need for the product rule on the right-hand side. Here, \( f(x) = x \) and \( g(x) = \ln(3x^2 + 2) \).

Applying the product rule, we get:

• The derivative of \( x \), which is 1.
• The derivative of \( \ln(3x^2 + 2) \), which involves the chain rule, resulting in \( \frac{1}{3x^2 + 2} \cdot 6x \).
• The complete derivative outcome includes both parts \( \ln(3x^2 + 2) \cdot 1 \) and \( x \cdot \frac{1}{3x^2 + 2} \cdot 6x \).

With this, you can understand how the product rule effectively aids in managing differentiation tasks when functions are multiplied together.

Derivative of Exponential Functions

The derivative of an exponential function is fundamental in calculus, frequently appearing in problems like our initial function \((3x^2 + 2)^x\). An exponential function generally takes the form \( a^x \), with a main rule where its derivative with respect to \( x \) is \( a^x \ln(a) \).

However, our example is slightly more complex due to the base \( 3x^2 + 2 \) that changes with \( x \) rather than being a constant. This complexity can be managed by using logarithms and implicit differentiation.

In our case:

• You start by taking the logarithm, turning it into \( x \ln(3x^2 + 2) \).
• Then apply differentiation rules, using both the product and chain rule, since the base is a function of \( x \).
• The result leads you to multiply back to return the derivative to its original non-log form, combining all derived components.

Understanding these rules helps in effectively handling exponential functions in calculus.

Implicit Differentiation

Implicit differentiation is a method used when a function cannot be easily solved for one variable in terms of another, making it tricky to differentiate directly. In exercises where functions are overlapping or nesting, employing implicit differentiation allows finding these derivatives indirectly.

By starting with \( \ln(y) = x \ln(3x^2 + 2) \), this strategy conveniently sidesteps direct complexity and allows working through derivative rules more smoothly:

• The derivative of \( \ln(y) \) with respect to \( x \) becomes \( \frac{y'}{y} \), highlighting the utility of implicit differentiation when \( y \) isn't isolated.
• This requires multiplying by the function value \( y \) after differentiating both sides, ensuring the derivative \( y' \) adheres to the function's relationship.

Implicit differentiation, thus, bridges gaps in direct calculus methods, making it a vital tactic in more intricate differentiation forms. This method maintains relationships through derivatives, ensuring complete solutions to complex calculus problems.