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

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:

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