`e^(mx).cos n x`

Answer

To find the differential coefficient of the function \( e^{mx} \cos(nx) \) with respect to \( x \), 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 \( uv \) is given by:

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

### Step-by-Step Solution:

1. **Identify the Functions**:
Let:
- \( u = e^{mx} \)
- \( v = \cos(nx) \)

2. **Differentiate \( u \) and \( v \)**:
- The derivative of \( u \) with respect to \( x \) is:
\[ \frac{du}{dx} = \frac{d}{dx}(e^{mx}) = m e^{mx} \]
- The derivative of \( v \) with respect to \( x \) is:
\[ \frac{dv}{dx} = \frac{d}{dx}(\cos(nx)) = -n \sin(nx) \]

3. **Apply the Product Rule**:
Now, applying the product rule:
\[ \frac{d}{dx}(e^{mx} \cos(nx)) = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ \frac{d}{dx}(e^{mx} \cos(nx)) = e^{mx} (-n \sin(nx)) + \cos(nx) (m e^{mx}) \]

4. **Simplify the Expression**:
Combine the terms:
\[ = -n e^{mx} \sin(nx) + m e^{mx} \cos(nx) \]
Factor out \( e^{mx} \):
\[ = e^{mx} (m \cos(nx) - n \sin(nx)) \]

### Final Answer:
Thus, the differential coefficient of the function \( e^{mx} \cos(nx) \) with respect to \( x \) is:
\[ \frac{d}{dx}(e^{mx} \cos(nx)) = e^{mx} (m \cos(nx) - n \sin(nx)) \]