Finding the tenth derivative of $f(x) = e^x\sin x$ at $x=0$

Hint:

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

Some details:

There results from the above remark and linearity of differentiation that $\;(\mathrm e^x\sin x)'=\bigl(\operatorname{Im}(\mathrm e^{(1+i)x})\bigr)'= \operatorname{Im}\bigl((1+i)\mathrm e^{(1+i)x}\bigr)$, hence $$\;(\mathrm e^x\sin x)''=\bigl(\operatorname{Im}((1+i)\mathrm e^{(1+i)x}))\bigr)'= \operatorname{Im}\bigl((1+i)^2\mathrm e^{(1+i)x}\bigr),$$ and more generally $$(\mathrm e^x\sin x)^{(k)}=\bigl(\operatorname{Im}(\mathrm e^{(1+i)x})\bigr)^{(k)}=\operatorname{Im}\bigl((1+i)^k(\mathrm e^{(1+i)x})\bigr).$$

Hint:

$$f(x)=e^x\sin x$$ $$f'(x)=e^x(\sin x +\cos x)$$ $$f''(x)=e^x(\sin x+\cos x)+e^x(\cos x -\sin x)=2e^x(\cos x)$$ $$f'''(x)=e^x(2\cos x)-e^x(2\sin x)=2e^x(\cos x-\sin x)$$ $$f^{IV}(x)=2e^x(\cos x-\sin x)-2e^x(\cos x+\sin x)=-4e^x(\sin x)=-4f(x)$$

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

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

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

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