If `y=cos^(-1) (sin x^(2)),"then " (dy)/(dx)=`

Options

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

Answer

To find the derivative of the function \( y = \cos^{-1}(\sin(x^2)) \), we will use the chain rule and the differentiation formula for the inverse cosine function.

### Step-by-Step Solution:

1. **Identify the outer and inner functions**:
- The outer function is \( y = \cos^{-1}(u) \) where \( u = \sin(x^2) \).
- The inner function is \( u = \sin(x^2) \).

2. **Differentiate the outer function**:
- The derivative of \( y = \cos^{-1}(u) \) is given by:
\[ \frac{dy}{du} = -\frac{1}{\sqrt{1 - u^2}} \]
- Substituting \( u = \sin(x^2) \):
\[ \frac{dy}{du} = -\frac{1}{\sqrt{1 - \sin^2(x^2)}} \]
- Using the identity \( 1 - \sin^2(x) = \cos^2(x) \):
\[ \frac{dy}{du} = -\frac{1}{\sqrt{\cos^2(x^2)}} = -\frac{1}{\cos(x^2)} \]

3. **Differentiate the inner function**:
- Now, we differentiate \( u = \sin(x^2) \):
\[ \frac{du}{dx} = \cos(x^2) \cdot \frac{d}{dx}(x^2) = \cos(x^2) \cdot 2x \]

4. **Apply the chain rule**:
- Now, we can apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
- Substituting the derivatives we found:
\[ \frac{dy}{dx} = -\frac{1}{\cos(x^2)} \cdot (2x \cos(x^2)) \]

5. **Simplify the expression**:
- The \( \cos(x^2) \) terms cancel out:
\[ \frac{dy}{dx} = -2x \]

### Final Answer:
\[ \frac{dy}{dx} = -2x \]