$f(x) = 2\sin 3x\cos 5x\\ f'(x)=(2\sin 3x)'\cos 5x+2\sin 3x(\cos 5x)'\\ =6\cos 3x\cos 5x-10\sin 3x\sin 5x\\ =6.\dfrac{1}{2}\left(\cos 8x+\cos2x\right)+10.\dfrac{1}{2}\left(\cos 8x-\cos2x\right)\\ =8\cos 8x-2\cos 2x\\ f'\left(\dfrac{\pi}{8}\right)\\ =8\cos \left(8.\dfrac{\pi}{8}\right)-2\cos \left(\dfrac{2.\pi}{8}\right)\\ =-8-\sqrt{2}$