Given $y = \sin(2x)$, to find the derivative $\frac{dy}{dx}$, we apply the chain rule.

The chain rule states that if $y = f(g(x))$, then:

$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

Here, $f(u) = \sin(u)$ and $g(x) = 2x$, so applying the chain rule:

1. The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$, so $f'(g(x)) = \cos(2x)$.
2. The derivative of $2x$ with respect to $x$ is 2.

Therefore, combining these:

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

So, $\frac{dy}{dx} = 2\cos(2x)$.

Would you like a more detailed explanation or any clarifications?

5 Related Questions:

1. How do you differentiate $y = \cos(3x)$?
2. What is the second derivative of $y = \sin(2x)$?
3. How does the chain rule apply in multi-variable calculus?
4. What is the derivative of $y = \sin^2(x)$?
5. Can you explain how to find the derivative of trigonometric functions?

Tip:

Always remember to check whether you need to apply the chain rule when differentiating composite functions.