Derivative of `cos x^3 sin ^2(x^5)` with respect to x is :

Options

A. `10x^4("sin"x^5) (cos x^5)(cos x^3)-3x^2 sin x^2 sin^2 x^5`
B. `5x^3(sinx^5)(cos x^5) (cos x^3) -3x^2 sin x^3 sin^2 x^5`
C. `5x^3("sin"x^5)(cos x^5)(cos x^3)-6x^2 sin x^3 sin^2 x^5`
D. None of these

Answer

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

### Step-by-Step Solution:

1. **Identify the Functions**:
Let \( u = \cos(x^3) \) and \( v = \sin^2(x^5) \). We need to find \( \frac{dy}{dx} \) where \( y = u \cdot v \).

2. **Apply the Product Rule**:
The product rule states that \( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \).

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

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

5. **Combine Using the Product Rule**:
Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) 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**:
Combine the terms:
\[ \frac{dy}{dx} = 10x^4 \cos(x^3) \sin(x^5) \cos(x^5) - 3x^2 \sin^2(x^5) \sin(x^3) \]

Thus, the derivative of \( \cos(x^3) \sin^2(x^5) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = 10x^4 \cos(x^3) \sin(x^5) \cos(x^5) - 3x^2 \sin^2(x^5) \sin(x^3) \]