To find the derivative of the function $y = \frac{\sin(2x)}{x^2}$, we need to use the quotient rule. The quotient rule states that if we have a function of the form $\frac{u(x)}{v(x)}$, its derivative is given by:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}$

In this case:

• $u(x) = \sin(2x)$
• $v(x) = x^2$

We first need to compute the derivatives $u'(x)$ and $v'(x)$.

Step 1: Compute $u'(x)$

$u(x) = \sin(2x) \implies u'(x) = \cos(2x) \cdot 2 = 2 \cos(2x)$

Step 2: Compute $v'(x)$

$v(x) = x^2 \implies v'(x) = 2x$

Step 3: Apply the Quotient Rule

Now we apply the quotient rule:

$\frac{dy}{dx} = \frac{x^2 \cdot (2 \cos(2x)) - \sin(2x) \cdot (2x)}{(x^2)^2}$

Simplifying the expression:

$\frac{dy}{dx} = \frac{2x^2 \cos(2x) - 2x \sin(2x)}{x^4}$

Step 4: Simplify the Expression

We can factor out a $2x$ from the numerator:

$\frac{dy}{dx} = \frac{2x (\cos(2x) - \frac{\sin(2x)}{x})}{x^4}$

Further simplifying:

$\frac{dy}{dx} = \frac{2 (\cos(2x) - \frac{\sin(2x)}{x})}{x^3}$

This is the derivative of the given function $y = \frac{\sin(2x)}{x^2}$.

Would you like more details on the steps? Feel free to ask any questions!

Here are 5 related questions that could expand your understanding:

1. How does the quotient rule differ from the product rule in calculus?
2. What are some common mistakes to avoid when using the quotient rule?
3. How can we apply L'Hopital's Rule to a function like $\frac{\sin(2x)}{x^2}$ as $x \to 0$?
4. What are the graphical interpretations of the derivative $\frac{dy}{dx}$?
5. Can we use trigonometric identities to simplify $\sin(2x)$ or $\cos(2x)$ in differentiation problems?

Tip: When differentiating trigonometric functions, keep in mind the chain rule to handle composed functions effectively!