If ` y= sin (cos (tan x)) ,then (dy)/(dx) =`

Options

A. ` -sin (cos (tan x )) sin (tan x ) sec ^(2) x `
B. ` sin (cos (tanx ))sin (tan x )sec ^(2) x`
C. ` -cos (cos (tan x ) )sin (tan x ) sec ^(2) x`
D. ` cos (cos (tan x ) )sin (tan x ) sec ^(2) x `

Answer

To find the derivative of the function \( y = \sin(\cos(\tan x)) \), we will use the chain rule of differentiation. The chain rule states that if you have a composite function, the derivative can be found by multiplying the derivatives of the outer function and the inner function.

### Step-by-Step Solution:

1. **Identify the outer and inner functions**:
- Let \( u = \cos(\tan x) \) (inner function)
- Therefore, \( y = \sin(u) \) (outer function)

2. **Differentiate the outer function**:
- The derivative of \( y = \sin(u) \) with respect to \( u \) is:
\[ \frac{dy}{du} = \cos(u) \]

3. **Differentiate the inner function**:
- Now we need to differentiate \( u = \cos(\tan x) \) with respect to \( x \).
- Let \( v = \tan x \) (inner function of \( u \))
- Therefore, \( u = \cos(v) \)
- The derivative of \( u = \cos(v) \) with respect to \( v \) is:
\[ \frac{du}{dv} = -\sin(v) \]

4. **Differentiate \( v = \tan x \)**:
- The derivative of \( v = \tan x \) with respect to \( x \) is:
\[ \frac{dv}{dx} = \sec^2(x) \]

5. **Apply the chain rule**:
- Now, using the chain rule:
\[ \frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = -\sin(v) \cdot \sec^2(x) \]
- Substitute back \( v = \tan x \):
\[ \frac{du}{dx} = -\sin(\tan x) \cdot \sec^2(x) \]

6. **Combine the derivatives**:
- Now we can find \( \frac{dy}{dx} \) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(u) \cdot \frac{du}{dx} \]
- Substitute back \( u = \cos(\tan x) \):
\[ \frac{dy}{dx} = \cos(\cos(\tan x)) \cdot \left(-\sin(\tan x) \cdot \sec^2(x)\right) \]

7. **Final expression**:
- Thus, the final answer is:
\[ \frac{dy}{dx} = -\cos(\cos(\tan x)) \cdot \sin(\tan x) \cdot \sec^2(x) \]