Question

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

Short answer

The derivative is \(f'(x) = 2x \ln(9x) + x\).

Step-by-step solution

Recognize the Product Rule

The function to differentiate, \(f(x) = x^2 \ln(9x)\), is a product of two functions: \(u(x) = x^2\) and \(v(x) = \ln(9x)\). Use the product rule, which states \((uv)' = u'v + uv'\).

Differentiate \(u(x) = x^2\)

Differentiate \(u(x) = x^2\). The derivative is \(u'(x) = 2x\).

Differentiate \(v(x) = \ln(9x)\) Using Chain Rule

Differentiate \(v(x) = \ln(9x)\). First, rewrite it as \(\ln(9) + \ln(x)\). The derivative of \(v(x)\) is \(v'(x) = \frac{1}{x}\) because \(\ln(9)\) is constant and does not affect differentiation.

Apply the Product Rule

Now apply the product rule: \(f'(x) = u'(x)v(x) + u(x)v'(x)\). Substitute \(u'(x) = 2x\), \(v(x) = \ln(9x)\), \(u(x) = x^2\), and \(v'(x) = \frac{1}{x}\) into the formula.

Formulate \(f'(x)\)

Substitute into the product rule: \[f'(x) = (2x)(\ln(9x)) + (x^2)\left(\frac{1}{x}\right)\].Simplify to get \[f'(x) = 2x \ln(9x) + x.\]

Simplify the Expression

Rewrite \(f'(x)\) as simply \(f'(x) = 2x \ln(9x) + x\). This is the derivative of the original function.

Key concepts

Product Rule

When dealing with differentiation, the Product Rule is a handy formula used to find the derivative of two multiplied functions. Simply put, if you're trying to differentiate a function that is the product of two smaller functions, the Product Rule comes to the rescue. For functions \(u(x)\) and \(v(x)\), the Product Rule is:

• \((uv)' = u'v + uv'\)

To use the Product Rule, follow these steps:

• First, differentiate \(u(x)\) to find \(u'(x)\).
• Next, differentiate \(v(x)\) to get \(v'(x)\).
• Finally, plug these into the formula \((uv)' = u'v + uv'\).

By systematically using the Product Rule, you can break down more complex differentiation problems into smaller, more manageable steps.

Chain Rule

The Chain Rule is vital when dealing with composite functions, or functions within other functions. If you have a function \(y = f(g(x))\), where \(g(x)\) is nested inside \(f(x)\), you will need the Chain Rule for differentiation. The Chain Rule says:

• \(\frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\)

To apply it:

• Differential the outer function \(f\) with respect to the inside function \(g\).
• Then, differentiate \(g(x)\) with respect to \(x\).
• Finally, multiply the two derivatives together.

This rule helps simplify the process by breaking it down.

Derivative

The concept of a derivative is fundamental in calculus. Derivatives represent the rate of change of a function as it relates to one of its variables. So, if you have a function \(f(x)\), the derivative, denoted as \(f'(x)\), shows how \(f\) changes with \(x\). The derivative of basic functions follows these rules:

• The derivative of \(x^n\) (where n is a constant) is \(nx^{n-1}\).
• The derivative of a constant is 0 because constants don’t change.

Finding derivatives requires practice but soon becomes a mechanical process.

Logarithmic Differentiation

Logarithmic Differentiation is a powerful technique especially useful for functions that involve logarithms, products, or even more complex expressions. This method uses properties of logarithms to simplify the differentiation process.
Think of a function \(y = x^x\). By applying logarithms, you start by taking the natural log of both sides, transforming it into \(\ln y = x \ln x\). This makes the differentiation easier.

• Differential each side of the log expression.
• With implicit differentiation, \(\frac{1}{y}\frac{dy}{dx} = \ln x + 1\).
• Then solve for \(\frac{dy}{dx}\), sometimes replacing \(y\) back with the original function.

This technique helps manage otherwise complicated or cumbersome expressions.