If `y=(e^(2x)cosx)/(x sinx),then(dy)/(dx)`=

Options

A. `(e^(2x)[(2x-1)cotx-xcosec^2x])/x^2`
B. `(e^(2x)[(2x+1)cotx-xcosec^2x])/x^2`
C. `(e^(2x)[(2x-1)cotx+xcosec^2x])/x^2`
D. None of these

Answer

To find the derivative of the function \( y = \frac{e^{2x} \cos x}{x \sin x} \), we will use the quotient rule of differentiation. The quotient rule states that if \( y = \frac{u}{v} \), then

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = e^{2x} \cos x \) and \( v = x \sin x \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = e^{2x} \cos x \)
- \( v = x \sin x \)

### Step 2: Differentiate \( u \) and \( v \)
Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

**Differentiating \( u \):**
Using the product rule \( \frac{d(uv)}{dx} = u'v + uv' \):
- \( u' = \frac{d}{dx}(e^{2x}) \cdot \cos x + e^{2x} \cdot \frac{d}{dx}(\cos x) \)
- \( u' = 2e^{2x} \cos x - e^{2x} \sin x \)
- Thus, \( \frac{du}{dx} = e^{2x} (2 \cos x - \sin x) \)

**Differentiating \( v \):**
Using the product rule again:
- \( v' = \frac{d}{dx}(x) \cdot \sin x + x \cdot \frac{d}{dx}(\sin x) \)
- \( v' = \sin x + x \cos x \)
- Thus, \( \frac{dv}{dx} = \sin x + x \cos x \)

### Step 3: Apply the Quotient Rule
Now we can apply the quotient rule:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Substituting the values we found:
\[ \frac{dy}{dx} = \frac{(x \sin x)(e^{2x} (2 \cos x - \sin x)) - (e^{2x} \cos x)(\sin x + x \cos x)}{(x \sin x)^2} \]

### Step 4: Simplify the Expression
Now, we simplify the numerator:
1. Expand the first term:
\[ x \sin x \cdot e^{2x} (2 \cos x - \sin x) = e^{2x} x \sin x (2 \cos x - \sin x) \]
2. Expand the second term:
\[ e^{2x} \cos x (\sin x + x \cos x) = e^{2x} \cos x \sin x + e^{2x} x \cos^2 x \]
3. Combine:
\[ \frac{dy}{dx} = \frac{e^{2x} \left[ x \sin x (2 \cos x - \sin x) - (\cos x \sin x + x \cos^2 x) \right]}{(x \sin x)^2} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{e^{2x} \left[ x \sin x (2 \cos x - \sin x) - (\cos x \sin x + x \cos^2 x) \right]}{(x \sin x)^2} \]