If `y=e^(x) sinx` then calculate `(dy)/(dx)`

Answer

To find the derivative of the function \( y = e^x \sin x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

In our case, we can identify:
- \( u = e^x \)
- \( v = \sin x \)

Now, we will differentiate both \( u \) and \( v \):

1. **Differentiate \( u = e^x \)**:
\[ \frac{du}{dx} = e^x \]

2. **Differentiate \( v = \sin x \)**:
\[ \frac{dv}{dx} = \cos x \]

Now, applying the product rule:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):

\[ \frac{dy}{dx} = e^x \cdot \cos x + \sin x \cdot e^x \]

This can be simplified to:

\[ \frac{dy}{dx} = e^x \cos x + e^x \sin x \]

Factoring out \( e^x \):

\[ \frac{dy}{dx} = e^x (\cos x + \sin x) \]

Thus, the final result for the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = e^x (\cos x + \sin x) \]