Derivative of y=sin(2x-1)*e^5x

1. Apply the product rule:

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

   $f{\left(x \right)} = e^{5} \sin{\left(2 x - 1 \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

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

      1. Let $u = 2 x - 1$.

      2. The derivative of sine is cosine:

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

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(2 x - 1\right)$:

         1. Differentiate $2 x - 1$ term by term:

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

               1. Apply the power rule: $x$ goes to $1$

               So, the result is: $2$

            2. The derivative of the constant $-1$ is zero.

            The result is: $2$

         The result of the chain rule is:

         $$2 \cos{\left(2 x - 1 \right)}$$

      So, the result is: $2 e^{5} \cos{\left(2 x - 1 \right)}$

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

   1. Apply the power rule: $x$ goes to $1$

   The result is: $2 x e^{5} \cos{\left(2 x - 1 \right)} + e^{5} \sin{\left(2 x - 1 \right)}$

2. Now simplify:

   $$\left(2 x \cos{\left(2 x - 1 \right)} + \sin{\left(2 x - 1 \right)}\right) e^{5}$$

The answer is:

$$\left(2 x \cos{\left(2 x - 1 \right)} + \sin{\left(2 x - 1 \right)}\right) e^{5}$$