To find the derivative of the function $h(x) = |3x| \cos(5x)$, we'll need to account for the fact that $|3x|$ involves the absolute value of $3x$. Let's break this down into parts:

Step 1: Break Down the Function

The function $h(x) = |3x| \cos(5x)$ can be split into two components:

• $|3x|$, which is the absolute value of $3x$,
• $\cos(5x)$, a trigonometric function.

We will use the product rule to differentiate, which states: $(fg)' = f'g + fg'$

Here, $f(x) = |3x|$ and $g(x) = \cos(5x)$.

Step 2: Derivative of $f(x) = |3x|$

The derivative of the absolute value function $|3x|$ is defined piecewise:

\begin{array}{ll} 1, & \text{if } x > 0 \\ -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \end{array} \right.$$ where $$\text{sgn}(x)$$ is the sign function. ### Step 3: Derivative of $$g(x) = \cos(5x)$$ Using the chain rule: $$\frac{d}{dx} \cos(5x) = -5 \sin(5x)$$ ### Step 4: Apply the Product Rule Using the product rule, we now compute $$h'(x)$$: $$h'(x) = \frac{d}{dx} (|3x|) \cdot \cos(5x) + |3x| \cdot \frac{d}{dx} (\cos(5x))$$ Substitute the derivatives: $$h'(x) = 3 \cdot \text{sgn}(x) \cdot \cos(5x) + |3x| \cdot (-5 \sin(5x))$$ Thus, the derivative is: $$h'(x) = 3 \cdot \text{sgn}(x) \cdot \cos(5x) - 5 |3x| \sin(5x)$$ ### Final Answer: $$h'(x) = 3 \cdot \text{sgn}(x) \cdot \cos(5x) - 15 |x| \sin(5x)$$ Would you like further details or clarifications? Here are 5 related questions: 1. How do we handle derivatives of absolute value functions in general? 2. What is the product rule, and how is it used? 3. Why do we need the sign function for differentiating absolute values? 4. Can we graph the original function and its derivative to visualize them? 5. What would happen if we change the cosine function to sine? **Tip:** When differentiating absolute value functions, remember to break them into piecewise definitions to avoid errors.