Derivative of \( \displaystyle - \frac{\sqrt{2} e^{2 x} \cos{\left(2 x + \frac{\pi}{4} \right)}}{4} \)

Problem 2.133 · hard

1. \[ \frac{d}{d x} \left(- \frac{\sqrt{2} e^{2 x} \cos{\left(2 x + \frac{\pi}{4} \right)}}{4}\right) \]
   derivativeStart with the derivative of the function.✓ Proved
2. \[ = - \frac{\sqrt{2} \frac{d}{d x} e^{2 x} \cos{\left(2 x + \frac{\pi}{4} \right)}}{4} \]
   constant-multiplePull out the constant factor.✓ Proved
3. \[ = - \frac{\sqrt{2} \left(e^{2 x} \frac{d}{d x} \cos{\left(2 x + \frac{\pi}{4} \right)} + \cos{\left(2 x + \frac{\pi}{4} \right)} \frac{d}{d x} e^{2 x}\right)}{4} \]
   productApply the product rule.✓ Proved
4. \[ = - \frac{\sqrt{2} \left(2 e^{2 x} \cos{\left(2 x + \frac{\pi}{4} \right)} + e^{2 x} \frac{d}{d x} \cos{\left(2 x + \frac{\pi}{4} \right)}\right)}{4} \]
   derivativeDifferentiate the first part of the product.✓ Proved
5. \[ = - \frac{\sqrt{2} \left(- 2 e^{2 x} \sin{\left(2 x + \frac{\pi}{4} \right)} + 2 e^{2 x} \cos{\left(2 x + \frac{\pi}{4} \right)}\right)}{4} \]
   chain algebraApply the chain rule to the cosine term. Simplify the signs.✓ Proved
6. \[ = - \frac{\sqrt{2} \left(- e^{2 x} \sin{\left(2 x + \frac{\pi}{4} \right)} + e^{2 x} \cos{\left(2 x + \frac{\pi}{4} \right)}\right)}{2} \]
   algebra algebraFactor out the common factor of 2. Simplify the constant coefficient.✓ Proved
7. \[ = - \frac{\sqrt{2} \left(- \sin{\left(2 x + \frac{\pi}{4} \right)} + \cos{\left(2 x + \frac{\pi}{4} \right)}\right) e^{2 x}}{2} \]
   algebraFactor out the exponential term.✓ Proved
8. \[ = e^{2 x} \sin{\left(2 x \right)} \]
   simplify simplifyUse the trigonometric identity cos(theta + pi/4) - sin(theta + pi/4) = -sqrt(2)sin(theta) where theta = 2x. Multiply the remaining terms to get the final simplified expression.✓ Proved