If `y=sin[cos(x^(2))]` then `(dy)/(dx)` is equal to

Options

A. `-2xsinx^(2).cos(cosx^(2))`
B. `-2xsinx^(2).cosx^(2)`
C. `2x^(2)sinx^(2).cosx^(2)`
D. None of these

Answer

To find the derivative of the function \( y = \sin(\cos(x^2)) \), we will apply the chain rule of differentiation. Here’s the step-by-step solution:

### Step 1: Identify the outer and inner functions
The function \( y \) can be expressed as a composition of functions:
- Outer function: \( u = \sin(v) \)
- Inner function: \( v = \cos(w) \) where \( w = x^2 \)

### Step 2: Differentiate the outer function
Using the chain rule, the derivative of \( y \) with respect to \( v \) is:
\[ \frac{dy}{dv} = \cos(v) \]

### Step 3: Differentiate the middle function
Next, we differentiate \( v = \cos(w) \) with respect to \( w \):
\[ \frac{dv}{dw} = -\sin(w) \]

### Step 4: Differentiate the innermost function
Now, we differentiate \( w = x^2 \) with respect to \( x \):
\[ \frac{dw}{dx} = 2x \]

### Step 5: Apply the chain rule
Now we can combine these derivatives using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot (2x) \]

### Step 6: Simplify the expression
Thus, we can write:
\[ \frac{dy}{dx} = -2x \sin(x^2) \cos(\cos(x^2)) \]

### Final Answer
The derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -2x \sin(x^2) \cos(\cos(x^2)) \]

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