$100$-th derivative of the function $f(x)=e^{x}\cos(x)$

HINT:

$e^x\cos x$ is the real part of $y=e^{(1+i)x}$

As $1+i=\sqrt2e^{i\pi/4}$

$y_n=(1+i)^ne^{(1+i)x}=2^{n/2}e^x\cdot e^{i(n\pi/4+x)}$

Can you take it from here?

Find fewer order derivatives:

\begin{align} f'(x)&=&e^x (\cos x -\sin x)&\longleftarrow&\\ f''(x)&=&e^x(\cos x -\sin x -\sin x -\cos x) \\ &=& -2e^x\sin x&\longleftarrow&\\ f'''(x)&=&-2e^x(\sin x + \cos x)&\longleftarrow&\\ f''''(x)&=& -2e^x(\sin x + \cos x + \cos x -\sin x)\\ &=& -4e^x \cos x \\ &=& -4f(x)&\longleftarrow&\\ &...&\\ \therefore f^{(100)}(\pi)&=&-4^{25} f(\pi) \end{align}

There is this more systematic approach (requires linear algebra) which we can extend for more complicated cases. The key fact is that you have a set of functions (vectors) whose linear span is closed under the derivative operator. (You will never get something that is not a linear combination of $e^x \cos(x)$ and $e^x \sin(x)$ by taking derivatives).

Consider the vector space $V$ generated by $e^x \cos(x), e^x \sin(x)$. The derivative $D:V\to V$ is a linear map in that space. In those basis vectors

$$D = \left( \begin{array}{cc} 1 & -1 \\ 1 & 1 \\ \end{array} \right)$$

The problem is now to calculate $D^{100}$. For that, we could diagonalise $D$. However, in this case, $D^4 = -4 I$ so $D^{100} = -4^{25} D$. Therefore

$$D(e_1) = 4 e_1 = e^x \cos(x)$$