if `y=x^2sinx+(3x)/(tanx)`, then `(dy)/(dx)` will be

Options

A. `2xsinx+x^2cosx+(3tanx-3xsec^2x)/(tan^2x)`
B. `2xsinx+(3xsec^2x-3tanx)/(tan^2x)`
C. `(x^2cosx+2xsinx-(3tanx-3xsec^2x))/(tan^2x)`
D. `(x^2cosx-2xsinx-(3tanx-3xsec^2x))/(tan^2)`

Explanation

To find the derivative of the function \( y = x^2 \sin x + \frac{3x}{\tan x} \), we will differentiate each term separately using the rules of differentiation.

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = x^2 \sin x + \frac{3x}{\tan x} \]

2. **Differentiate the first term \( x^2 \sin x \)**:
- We will use the product rule here, which states that if \( u = x^2 \) and \( v = \sin x \), then:
\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]
- Here, \( u = x^2 \) and \( v = \sin x \):
- \( \frac{du}{dx} = 2x \)
- \( \frac{dv}{dx} = \cos x \)
- Applying the product rule:
\[ \frac{d}{dx}(x^2 \sin x) = x^2 \cos x + \sin x \cdot 2x = x^2 \cos x + 2x \sin x \]

3. **Differentiate the second term \( \frac{3x}{\tan x} \)**:
- We will use the quotient rule here, which states that if \( u = 3x \) and \( v = \tan x \), then:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
- Here, \( u = 3x \) and \( v = \tan x \):
- \( \frac{du}{dx} = 3 \)
- \( \frac{dv}{dx} = \sec^2 x \)
- Applying the quotient rule:
\[ \frac{d}{dx}\left(\frac{3x}{\tan x}\right) = \frac{\tan x \cdot 3 - 3x \cdot \sec^2 x}{\tan^2 x} \]

4. **Combine the results**:
- Now, we can combine the derivatives from both terms:
\[ \frac{dy}{dx} = \left(x^2 \cos x + 2x \sin x\right) + \frac{3 \tan x - 3x \sec^2 x}{\tan^2 x} \]

5. **Final expression**:
- Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = x^2 \cos x + 2x \sin x + \frac{3 \tan x - 3x \sec^2 x}{\tan^2 x} \]