"Trig-ky" Derivative

$\begin{aligned} f(x) & = \sin^4 x + \cos^4 x \\ & = (\sin^2x+\cos^2 x)^2 - 2\sin^2 x \cos^2 x \\ & = 1 - \frac 12 \sin^2 (2x) \\ & = 1 - \frac 14(1-\cos (4x)) \\ & = \frac 34 + \frac 14 \color{#3D99F6} \cos (4x) & \small \color{#3D99F6} \text{By Euler's formula }e^{\theta i} = \cos \theta + i \sin \theta \\ & = \frac 34 + \frac 14 \color{#3D99F6} \Re \left(e^{4xi}\right) & \small \color{#3D99F6} \text{where } \Re(z) \text{ denotes the real part of }z. \end{aligned}$

Then we have:

$\begin{aligned} f'(x) & = \frac d{dx} \left(\frac 34 + \frac 14 \Re \left(e^{4xi}\right) \right) = \Re \left(i e^{4xi}\right) \\ f'' (x) & = \Re \left(4i^2 e^{4xi}\right) \\ f''' (x) & = \Re \left(4^2i^3 e^{4xi}\right) \\ \implies f^{(n)} (x) & = \Re \left(4^{(n-1)}i^n e^{4xi}\right) & \small \color{#3D99F6} \text{where } f^{(n)} \text{ denotes }\frac {d^n}{dx^n}. \end{aligned}$

Therefore,

$\begin{aligned} f^{(2019)} \left(-\frac {5\pi}{16} \right) & = \Re \left(4^{2018}i^{2019} e^{-\frac 54\pi i}\right) \\ & = \Re \left(2^{4036}(-i) \left(\cos \frac 54 \pi - i \sin \frac 54 \pi \right) \right) \\ & = - 2^{4036} \sin \frac 54 \pi = 2^{4036} \times \frac 1{\sqrt 2} \\ & = 2^{4035.5} \end{aligned}$

Implying $a+b= 2+4035.5 = \boxed{4037.5}$ .

After some simplification the function we have $f(x) = 1 -\frac{\sin^22x}{2}=1-\frac{1}{4}-\frac{\cos4x}{4}$ Since the derivatives of constants vanishes. Left to derivate $\frac{\cos 4x}{4}$ . As $f^n(\cos (ax+b) )=a^n\cos \left(ax+b+\frac{n\pi}{2}\right)$ Set $a=4 ,b=0$ multiple by $-\frac{1}{4}$ $-\frac{1}{4}f^n\cos4x =-\dfrac{4^n}{4}\cos\left(4x+\frac{n\pi}{2}\right)=-4^{n-1}\cos\left(4x+\frac{n\pi}{2}\right)$ Thus required derivates at $n=2019$ and $x=-\frac{5\pi}{16}$ is $=-4^{2018}\cos\left(-\frac{5\pi}{4}+\frac{2019\pi}{2}\right)=-4^{2018} \cos\left(\frac{5\pi}{4}\right)=\dfrac{2^{4036}}{\sqrt 2}=2^{4036-0.5}$ And gives $a+b=4037.5$