$100$-th derivative of the function $f(x)=e^{x}\cos(x)$ I've got this task I'm not able to solve. So i need to find the 100-th derivative of $$f(x)=e^{x}\cos(x)$$ where $x=\pi$.
I've tried using Leibniz's formula but it got me nowhere, induction doesn't seem to help either, so if you could just give me a hint, I'd be very grateful.
Many thanks!
 A: 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
$$D(e_1) = 4 e_1 = e^x \cos(x)$$
A: Alternatively, one may use the General Leibniz rule/Cauchy formula
$$
(fg)^{(n)}(x)=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}(x)g^{(k)}(x)
$$ with
$$
f(x)=e^x,\quad f^{(n-k)}(x)=e^x,\quad g(x)=\cos x,\quad g^{(k)}(x)=\cos(x+k\pi/2).
$$
A: There is a pattern, which is easy to see in this problem,
(e^x = @)
@cosx
1st derivative - @cosx -@sinx
2nd derivative - -2@sinx
3rd ..         - -2(@sinx + @cosx)
4th ..         - -4@cosx
100 = 0mod4
So, 100th derivative of @cosx should be of the form a@cosx and a is (-4)^25.
So, at x = π, 100th derivative is 4²⁵e^π.
A: I'm a little late to the game, but you can check your work in Wolfram Alpha with this query:
D[E^x Cos[x], {x, 100}]  at x = pi

which matches the answer posted by @choco_addicted. The syntax comes from the Wolfram Language page for the D function.
A: $\cos(x) = \frac {\exp(\mathrm{i}x) + \exp (-\mathrm{i}x)}2$
And the rest is basically grunt work.
A: 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?
A: 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}
