If `y=sin^(-1) (cos 3x),"then "dy/dx=`

Options

A. ` 3`
B. ` -3`
C. ` (3)/(sqrt( 1-x^(2)))`
D. ` (-3)/(sqrt( 1-x^(2)))`

Answer

To solve the problem where \( y = \sin^{-1}(\cos(3x)) \) and we need to find \( \frac{dy}{dx} \), we can follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)
We start by applying the chain rule of differentiation. The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is given by:
\[ \frac{d}{du}(\sin^{-1}(u)) = \frac{1}{\sqrt{1 - u^2}} \]
In our case, \( u = \cos(3x) \).

### Step 2: Apply the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (\cos(3x))^2}} \cdot \frac{d}{dx}(\cos(3x)) \]

### Step 3: Differentiate \( \cos(3x) \)
Now we need to differentiate \( \cos(3x) \):
\[ \frac{d}{dx}(\cos(3x)) = -\sin(3x) \cdot \frac{d}{dx}(3x) = -\sin(3x) \cdot 3 = -3\sin(3x) \]

### Step 4: Substitute back into the derivative
Now we substitute this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (\cos(3x))^2}} \cdot (-3\sin(3x)) \]

### Step 5: Simplify the expression
Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we know:
\[ 1 - \cos^2(3x) = \sin^2(3x) \]
Thus, we can rewrite the square root:
\[ \sqrt{1 - \cos^2(3x)} = \sqrt{\sin^2(3x)} = |\sin(3x)| \]
Assuming \( \sin(3x) \) is non-negative in the interval we are considering, we can simplify:
\[ \frac{dy}{dx} = \frac{-3\sin(3x)}{|\sin(3x)|} \]
This simplifies to:
\[ \frac{dy}{dx} = -3 \quad \text{(if } \sin(3x) \text{ is positive)} \]

### Final Answer
Thus, the final answer is:
\[ \frac{dy}{dx} = -3 \]