Find the nth differential coefficient of sin 2x.

Steps to Solve

1. Understanding the nth Derivative Concept
   To find the nth differential coefficient of a function, we need to identify a general formula or method for taking derivatives multiple times. For periodic functions like $\sin(2x)$, this can often be simplified.

2. Finding the First Derivative
   The first derivative of $\sin(2x)$ can be found using the chain rule:
   $$ \frac{d}{dx}[\sin(2x)] = 2 \cos(2x) $$

3. Finding the Second Derivative
   Now, calculate the second derivative:
   $$ \frac{d^2}{dx^2}[\sin(2x)] = \frac{d}{dx}[2 \cos(2x)] = -4 \sin(2x) $$

4. Continuing the Pattern
   The third derivative is:
   $$ \frac{d^3}{dx^3}[\sin(2x)] = \frac{d}{dx}[-4 \sin(2x)] = -8 \cos(2x) $$
   The fourth derivative is:
   $$ \frac{d^4}{dx^4}[\sin(2x)] = \frac{d}{dx}[-8 \cos(2x)] = 16 \sin(2x) $$

5. Identifying the nth Derivative Pattern
   From the derivatives calculated, we can see a pattern:

• For $n = 1$: $2 \cos(2x)$
• For $n = 2$: $-4 \sin(2x)$
• For $n = 3$: $-8 \cos(2x)$
• For $n = 4$: $16 \sin(2x)$

The pattern appears to alternate between $\sin(2x)$ and $\cos(2x)$, with coefficients that are powers of 2. For even $n$, the derivative has the form $2^n \sin(2x)$, and for odd $n$, it has the form $2^n \cos(2x)$.

6. Writing the General Formula
   We can summarize the nth derivative as follows:
   $$ \frac{d^n}{dx^n}[\sin(2x)] = 2^n \sin\left(2x + \frac{n\pi}{2}\right) $$
   This accounts for the alternating nature and scaling factor.