Derivative of e^x(sinx-cosx)

1. Apply the product rule:

   $$\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}$$

   $f{\left(x \right)} = e^{x}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. The derivative of $e^{x}$ is itself.

   $g{\left(x \right)} = \sin{\left(x \right)} - \cos{\left(x \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $\sin{\left(x \right)} - \cos{\left(x \right)}$ term by term:

      1. The derivative of sine is cosine:

         $$\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$$

      2. The derivative of a constant times a function is the constant times the derivative of the function.

         1. The derivative of cosine is negative sine:

            $$\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}$$

         So, the result is: $\sin{\left(x \right)}$

      The result is: $\sin{\left(x \right)} + \cos{\left(x \right)}$

   The result is: $\left(\sin{\left(x \right)} - \cos{\left(x \right)}\right) e^{x} + \left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{x}$

2. Now simplify:

   $$2 e^{x} \sin{\left(x \right)}$$

The answer is: