Derivative of е^sinx*sinx

1. Apply the product rule:

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

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

   1. Let $u = \sin{\left(x \right)}$.

   2. The derivative of $e^{u}$ is itself.

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \sin{\left(x \right)}$:

      1. The derivative of sine is cosine:

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

      The result of the chain rule is:

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

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

   1. The derivative of sine is cosine:

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

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

2. Now simplify:

   $$\left(\sin{\left(x \right)} + 1\right) e^{\sin{\left(x \right)}} \cos{\left(x \right)}$$

The answer is: