If `u=e^(ax)sinbx` and `v=e^(ax)cosbx`. then what is `u(du)/(dx)+v(dv)/(dx)` equal to ?

Options

A. A) `ae^(2ax)`
B. B) `(a^(2)+b^(2))e^(ax)`
C. C) `abe^(2ax)`
D. D) `(a+b)e^(ax)`

Answer

To solve the problem, we need to calculate \( u \frac{du}{dx} + v \frac{dv}{dx} \) where \( u = e^{ax} \sin(bx) \) and \( v = e^{ax} \cos(bx) \).

### Step-by-Step Solution:

1. **Find \( \frac{du}{dx} \)**:
- Using the product rule for derivatives, we have:
\[ \frac{du}{dx} = \frac{d}{dx}(e^{ax} \sin(bx)) = e^{ax} \cdot \frac{d}{dx}(\sin(bx)) + \sin(bx) \cdot \frac{d}{dx}(e^{ax}) \]
- The derivative of \( \sin(bx) \) is \( b \cos(bx) \) and the derivative of \( e^{ax} \) is \( ae^{ax} \).
- Thus,
\[ \frac{du}{dx} = e^{ax} (b \cos(bx)) + \sin(bx) (ae^{ax}) = e^{ax} (b \cos(bx) + a \sin(bx)) \]

2. **Find \( \frac{dv}{dx} \)**:
- Similarly, using the product rule:
\[ \frac{dv}{dx} = \frac{d}{dx}(e^{ax} \cos(bx)) = e^{ax} \cdot \frac{d}{dx}(\cos(bx)) + \cos(bx) \cdot \frac{d}{dx}(e^{ax}) \]
- The derivative of \( \cos(bx) \) is \( -b \sin(bx) \).
- Thus,
\[ \frac{dv}{dx} = e^{ax} (-b \sin(bx)) + \cos(bx) (ae^{ax}) = e^{ax} (-b \sin(bx) + a \cos(bx)) \]

3. **Calculate \( u \frac{du}{dx} + v \frac{dv}{dx} \)**:
- Substitute \( u \) and \( v \) into the expression:
\[ u \frac{du}{dx} = e^{ax} \sin(bx) \cdot e^{ax} (b \cos(bx) + a \sin(bx)) = e^{2ax} \sin(bx)(b \cos(bx) + a \sin(bx)) \]
\[ v \frac{dv}{dx} = e^{ax} \cos(bx) \cdot e^{ax} (-b \sin(bx) + a \cos(bx)) = e^{2ax} \cos(bx)(-b \sin(bx) + a \cos(bx)) \]
- Now combine the two results:
\[ u \frac{du}{dx} + v \frac{dv}{dx} = e^{2ax} \left( \sin(bx)(b \cos(bx) + a \sin(bx)) + \cos(bx)(-b \sin(bx) + a \cos(bx)) \right) \]

4. **Simplify the expression**:
- Distributing and combining like terms:
\[ = e^{2ax} \left( ab \sin^2(bx) + a \sin(bx) \cos(bx) + a \cos^2(bx) - b \sin(bx) \cos(bx) \right) \]
- This can be rearranged:
\[ = e^{2ax} \left( a(\sin^2(bx) + \cos^2(bx)) + (a - b) \sin(bx) \cos(bx) \right) \]
- Using the identity \( \sin^2(bx) + \cos^2(bx) = 1 \):
\[ = e^{2ax} \left( a + (a - b) \sin(bx) \cos(bx) \right) \]

### Final Result:
Thus, the final result for \( u \frac{du}{dx} + v \frac{dv}{dx} \) is:
\[ u \frac{du}{dx} + v \frac{dv}{dx} = e^{2ax} \left( a + (a - b) \sin(bx) \cos(bx) \right) \]