Question - Finding the Derivative of Sine Function with a Scalar Multiple of the Variable

To find the derivative of the function \( f(x) = \sin(2x) \), we will use the chain rule of differentiation. The chain rule is a formula to compute the derivative of a composite function. If we have a function \( h(x) = g(f(x)) \), then the chain rule states that:

\[ h'(x) = g'(f(x)) \cdot f'(x) \]

In your case, \( f(x) = \sin(2x) \) can be thought of as a composition of two functions: \( u(x) = 2x \) (the inner function) and \( \sin(u) \) (the outer function).

Let's differentiate \( u(x) = 2x \) first:

\[ u'(x) = \frac{d}{dx}(2x) = 2 \]

Next, we differentiate \( \sin(u) \) with respect to \( u \):

\[ \frac{d}{du}(\sin(u)) = \cos(u) \]

Now we apply the chain rule, which tells us to multiply the derivative of the outside function (with the inner function plugged in) by the derivative of the inside function:

\[ \frac{d}{dx}[\sin(2x)] = \cos(2x) \cdot \frac{d}{dx}(2x) \]

Putting it all together:

\[ \frac{d}{dx}[\sin(2x)] = \cos(2x) \cdot 2 \]

Therefore, the derivative of \( f(x) = \sin(2x) \) is:

\[ f'(x) = 2\cos(2x) \]

This completed derivative tells us the rate of change of \( f(x) \) with respect to \( x \). Remember, the factor of 2 comes from the derivative of \( 2x \), and the cosine comes from the derivative of \( \sin(u) \) with respect to \( u \).