Problem 342 - Calculus Volume 1

Question

Find \(f^{\prime}(x)\) for each function. $$ f(x)=x^{2} \ln 9 x $$

Step-by-Step Solution

Answer

The derivative is \(f^{\prime}(x) = x(2 \ln(9x) + 1)\).

Step 1: Differentiate using the Product Rule

To differentiate the product of two functions, we use the product rule. If we have a function given by \( f(x) = u(x) \cdot v(x) \), then the derivative \( f^{\prime}(x) \) is given by \( u^{\prime}(x) v(x) + u(x)v^{\prime}(x) \). In this exercise, let \( u(x) = x^2 \) and \( v(x) = \ln(9x) \).

Step 2: Differentiate u(x)

For \( u(x) = x^2 \), the derivative, using the basic power rule, is \( u^{\prime}(x) = 2x \).

Step 3: Differentiate v(x)

For \( v(x) = \ln(9x) \), we use the chain rule. The derivative of \( \ln(9x) \) is \( \frac{d}{dx}[\ln(9x)] = \frac{1}{9x} \times 9 = \frac{1}{x} \), since the derivative of \( \ln(x) \) is \( \frac{1}{x} \) and the constant factor of 9 comes from applying the chain rule.

Step 4: Apply the Product Rule

Using the product rule, substitute \( u^{\prime}(x) = 2x \), \( v(x) = \ln(9x) \), \( u(x) = x^2 \), and \( v^{\prime}(x) = \frac{1}{x} \):\[f^{\prime}(x) = 2x \cdot \ln(9x) + x^2 \cdot \frac{1}{x}.\]

Step 5: Simplify the Expression

Simplify the derivative from Step 4:\[2x \ln(9x) + x = x(2 \ln(9x) + 1).\]

Step 6: Final Expression for f'(x)

The simplified expression for the derivative is\[f^{\prime}(x) = x(2 \ln(9x) + 1).\]

Key Concepts

Product Rule

When you differentiate a product of two functions, the product rule is your best friend. It provides a formula for understanding how to differentiate products easily. Imagine you have a function where two smaller functions are multiplied together, like in the exercise: \(f(x) = u(x) \cdot v(x)\).

• For the product rule, identify both parts of the product: \(u(x)\) and \(v(x)\).
• Differentiate each part separately: \(u'(x)\) and \(v'(x)\).
• Combine them using the product rule formula: \(f'(x) = u'(x)v(x) + u(x)v'(x)\).

The key takeaway is to treat them in pieces. Calculate their derivatives individually, then put them together with the product rule's structure.

Power Rule

Differentiating polynomial expressions is straightforward with the power rule. This rule is especially handy for functions of the form \(x^n\), where "n" is any real number.

• The power rule states: "Bring down the exponent as a multiplier, then decrease the exponent by one."
• For example, if you have \(u(x) = x^2\), then \(u'(x) = 2x\).
• This simple rule helps quickly find derivatives of simple polynomial terms.

Understanding the power rule's simplicity and ease of application will drastically speed up working through problems involving differentiation of polynomial functions.

Chain Rule

The chain rule is a gateway for differentiating composite functions. Think of it as peeling an onion: you handle each layer, or function, separately.

• The chain rule tackles functions inside functions, like \(v(x) = \ln(9x)\).
• To apply the chain rule, find the derivative of the outer function while keeping the inner function untouched. Then multiply by the derivative of the inner function.
• For example, the derivative of \(\ln(9x)\) becomes \(\frac{1}{9x} \times 9 = \frac{1}{x}\).

This rule is intricate but methodical, making it invaluable for complex expressions where functions nest within one another. Knowing when and how to apply the chain rule is crucial for mastering calculus differentiation problems.