To find the derivative of $y = \ln(x) \sin(3x)$, we need to apply the product rule and the chain rule.

Product Rule:

The product rule states that if $y = u(x) \cdot v(x)$, then:

$\frac{dy}{dx} = u'(x) v(x) + u(x) v'(x)$

In our case:

• $u(x) = \ln(x)$
• $v(x) = \sin(3x)$

Step 1: Differentiate $u(x) = \ln(x)$

The derivative of $\ln(x)$ is: $u'(x) = \frac{1}{x}$

Step 2: Differentiate $v(x) = \sin(3x)$

The derivative of $\sin(3x)$ requires the chain rule. The derivative of $\sin(z)$ is $\cos(z)$, and we multiply by the derivative of the inner function $3x$, which is 3:

$v'(x) = 3 \cos(3x)$

Step 3: Apply the Product Rule

Now, using the product rule:

$\frac{dy}{dx} = u'(x) v(x) + u(x) v'(x)$

Substituting the derivatives:

$\frac{dy}{dx} = \frac{1}{x} \sin(3x) + \ln(x) \cdot 3 \cos(3x)$

Final Answer:

$\frac{dy}{dx} = \frac{\sin(3x)}{x} + 3 \ln(x) \cos(3x)$

Would you like further explanation or have any questions on the process?

Here are 5 related questions:

1. How do you apply the chain rule to differentiate trigonometric functions?
2. Can you derive the product rule from first principles?
3. What happens if we have an exponential function multiplied by a logarithmic function?
4. How would you handle the differentiation of $\ln(x^2 + 1)$?
5. Can you explain the derivative of $\cos(3x)$ with respect to $x$?

Tip: Always check if the function involves a product or a quotient before selecting the rule (product rule, quotient rule, chain rule, etc.).