`d/(dx)[sin {2 cos^(-1) (sin x)}]`=

Options

A. `-2 sin 2x`
B. `-2 cos 2x`
C. 2 sin 2x
D. 2 cos 2x

Answer

To solve the problem \( \frac{d}{dx} \left[ \sin \left( 2 \cos^{-1} (\sin x) \right) \right] \), we will follow these steps:

### Step 1: Set up the function
Let
\[ y = \sin \left( 2 \cos^{-1} (\sin x) \right) \]

### Step 2: Use the identity for cosine inverse
We know that
\[ \cos^{-1} (\sin x) = \frac{\pi}{2} - x \quad \text{(for } x \text{ in the range of } [0, \frac{\pi}{2})\text{)} \]
Thus, we can rewrite \(y\) as:
\[ y = \sin \left( 2 \left( \frac{\pi}{2} - x \right) \right) \]

### Step 3: Simplify the expression
Now simplify:
\[ y = \sin \left( \pi - 2x \right) \]
Using the property of sine, we have:
\[ \sin(\pi - \theta) = \sin(\theta) \]
So,
\[ y = \sin(2x) \]

### Step 4: Differentiate with respect to \(x\)
Now we differentiate \(y\):
\[ \frac{dy}{dx} = \frac{d}{dx} \left[ \sin(2x) \right] \]
Using the chain rule:
\[ \frac{dy}{dx} = 2 \cos(2x) \]

### Final Result
Thus, the derivative is:
\[ \frac{d}{dx} \left[ \sin \left( 2 \cos^{-1} (\sin x) \right) \right] = 2 \cos(2x) \]

---