If ` y=cos (x^(3)) sin ^(2) (x^(5)),then (dy)/(dx) =`

Options

A. ` 5x^(4)sin (x^(3))sin (2x^(5))-3x^(2) cos (x^(3))sin ( x^(5))`
B. ` 5x^(4)sin (x^(3))sin (2x^(5))+ 3x^(2) cos (x^(3))sin ( x^(5))`
C. ` 5x^(4)sin (x^(3))sin (2x^(5))- 3x^(2) sin (x^(3))sin^(2) ( x^(5))`
D. ` 5x^(4)sin (x^(3))sin (2x^(5))+ 3x^(2) sin (x^(3))sin^(2) ( x^(5))`

Answer

To find the derivative of the function \( y = \cos(x^3) \sin^2(x^5) \), we will use the product rule and the chain rule of differentiation.

### Step-by-Step Solution:

1. **Identify the components of the function**:
The function can be expressed as:
\[ y = u \cdot v \]
where \( u = \cos(x^3) \) and \( v = \sin^2(x^5) \).

2. **Apply the product rule**:
The product rule states that:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

3. **Find \( \frac{du}{dx} \)**:
To differentiate \( u = \cos(x^3) \), we will apply the chain rule:
\[ \frac{du}{dx} = -\sin(x^3) \cdot \frac{d}{dx}(x^3) = -\sin(x^3) \cdot 3x^2 \]
So,
\[ \frac{du}{dx} = -3x^2 \sin(x^3) \]

4. **Find \( \frac{dv}{dx} \)**:
To differentiate \( v = \sin^2(x^5) \), we again use the chain rule:
\[ \frac{dv}{dx} = 2\sin(x^5) \cdot \frac{d}{dx}(\sin(x^5)) = 2\sin(x^5) \cdot \cos(x^5) \cdot \frac{d}{dx}(x^5) = 2\sin(x^5) \cdot \cos(x^5) \cdot 5x^4 \]
So,
\[ \frac{dv}{dx} = 10x^4 \sin(x^5) \cos(x^5) \]

5. **Substitute back into the product rule**:
Now substituting \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) back into the product rule:
\[ \frac{dy}{dx} = \cos(x^3) \cdot (10x^4 \sin(x^5) \cos(x^5)) + \sin^2(x^5) \cdot (-3x^2 \sin(x^3)) \]

6. **Simplify the expression**:
\[ \frac{dy}{dx} = 10x^4 \cos(x^3) \sin(x^5) \cos(x^5) - 3x^2 \sin^2(x^5) \sin(x^3) \]

### Final Answer:
\[ \frac{dy}{dx} = 10x^4 \cos(x^3) \sin(x^5) \cos(x^5) - 3x^2 \sin^2(x^5) \sin(x^3) \]