Gfyuki

Hint:

As $;mathrm e^xsin x=operatorname{Im}bigl(mathrm e^{(1+i)x}bigr)$, you have to find first the real and imaginary parts of $(1+i)^{10}$.

Using power series: it is well-known that $e^x=sum_{k=0}^infty frac{x^k}{k!}$ and $sin(x)=sum_{k=0}^infty (-1)^kfrac{x^{2k+1}}{(2k+1)!}$ for any real number $x$, so

$$ e^xsin(x)=(1+x+frac{x^2}{2!}+dots+frac{x^{10}}{10!}+dots)(x-frac{x^3}{3!}+frac{x^5}{5!}-frac{x^7}{7!}+frac{x^9}{9!}+dots) $$

By expanding, the coefficient of $x^{10}$ is $frac{1}{9!1!}-frac{1}{7!3!}+frac{1}{5!5!}-frac{1}{7!3!}+frac{1}{9!1!}$

But this coefficient is also $frac{f^{(10)}(0)}{10!}$, so

$$ f^{(10)}(0)=frac{10!}{9!1!}-frac{10!}{7!3!}+frac{10!}{5!5!}-frac{10!}{7!3!}+frac{10!}{9!1!} = 10 -120 + 252 - 120 +10 = 32$$

Use Leibniz' Rule for higher derivatives of a product:
$$frac{d^n}{dx^n}(uv)
=frac{d^nu}{dx^n}v+binom n1frac{d^{n-1}u}{dx^{n-1}}frac{dv}{dx}
+binom n2frac{d^{n-2}u}{dx^{n-2}}frac{d^2v}{dx^2}+cdots+ufrac{d^nv}{dx^n} .$$
In your case take $u=e^x$ and $v=sin x$. Since you are going to substitute $x=0$ after differentiating, all the $e^x$ terms will be $1$, all the $sin x$ terms will be $0$ and all the cos $x$ terms will be $1$ (though some of them will pick up a negative sign when you differentiate). So the answer is
$$eqalign{0+binom{10}11&{}+binom{10}20+binom{10}3(-1)+binom{10}40+binom{10}51cr
&qquad{}+binom{10}60+binom{10}7(-1)+binom{10}80+binom{10}91+binom{10}{10}0cr
&=10-120+252-120+10cr
&=32 .cr}$$

$$(e^x(acos x+bsin x))'=e^x(acos x+bsin x)+e^x(bcos x-asin x)=e^x((a+b)cos x+(b-a)sin x).$$

So

$$(0,1)to(1,1)to(2,0)to(2,-2)to(0,-4)to(-4,-4)to(-8,0)to(-8,8)to(0,16)to(16,16)to(32,0).$$

If you divide by increasing powers of $2$, in pairs, the pattern emerges, with period $8$:

$$(0,1)to(1,1)to(1,0)to(1,-1)to(0,-1)to(-1,-1)to(-1,0)to(-1,1)to(0,1)to(1,1)to(1,0).$$

One trick here is to use $e^{ix}=cos x+isin x$ and define $g(x)=e^xcos x$ then $f(x)$ and $g(x)$ are both real functions.

Let $h(x)=g(x)+if(x)=e^{(1+i)x}$ then the tenth derivative of $h(x)$ is $(1+i)^{10}h(x)$ and the tenth derivative of $f(x)$ is the imaginary part of this.

Because you only want the value at $x=0$ you can evaluate there, with $h(0)=g(0)+if(0)$

Use the formula $(fg)^{(n)}= sumlimits_{k=0}^{n} binom {n} {k} (f)^{(k)}(g)^{(n-k)}$.

I get $f^{10}(x)=32e^xcos x$.

Here's what I did:
begin{align}f'(x)&=e^x(sin x+cos x)\
implies f''(x)&=2e^xcos x\
implies f^3(x)&=2e^x(cos x-sin x)\
implies f^4(x)&=2e^x(-2sin x)=-4f(x)\
implies f^5(x)&=-4f'(x)\
implies f^8(x)&=-4f^4(x)=16f(x)\
implies f^{10}(x)&=16f''(x)end{align}

Regarding the question about looking for a pattern:

Repeated application of the product rule gives

$$ f(x) = e^x sin x$$
$$ f'(x) = e^x sin x + e^x cos x$$
$$ f''(x) = e^x sin x + e^x cos x + e^x cos x - e^x sin x = 2 e^x cos x$$
Is a pattern emerging?
$$ f^{(3)}(x) = 2 e^x cos x -2 e^x sin x$$
$$ f^{(4)}(x) = 2 e^x cos x -2 e^x sin x - 2 e^x sin x - 2 e^x cos x = -4 e^x sin x$$
Yes. We can then conclude that

$ f^{(6)}(x) = -8 e^x cos x$, $f^{(8)}(x) = 16 e^x sin x$, and $f^{(10)}(x) = 32 e^x cos x$

such that $f^{(10)}(0) = 32 $