A basic question on successive differentiation How to prove that $$\frac{d^r}{dx^r}\cos x + i\frac{d^r}{dx^r}\sin x = i^r e^{ix}\ ?$$
I can understand it by putting values, but how to prove it? 
 A: $$\frac{d^r}{dx^r}\cos x  + i\frac{d^r}{dx^r} \sin x =\\=\frac{d^r}{dx^r}\left(\cos x + i \sin x\right) = \\ = \frac{d^r}{dx^r} e^{ix}$$
A: One way, using the ease to differentiate the exponential function, the definition of $\;e^{ix}\;,\;\;x\in\Bbb R\;$ and the linearity of the differentiation operators, I can think of is:
$$x\in\Bbb R\;,\;\;e^{ix}:=\cos x +i\sin x\implies$$
$$ i^re^{ix}= \frac{d^r}{dx^r}\left(e^{ix}\right)=\frac{d^r}{dx^r}\left(\cos x+i\sin x\right)=\frac{d^r}{dx^r}(\cos x)+i\frac{d^r}{dx^r}(\sin x)$$
A: Aside from directly observing your expression is equal to $\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}e^{ix}$, we know that $\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}\left(\cos(x)\right)$ and $\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}\left(\sin(x)\right)$ are cyclical modulo $4$, and therefore we only need to examine the case of $r\equiv\{0,1,2,3\} \pmod{4}$.
This gives us:


*

*For $r\equiv 0\pmod{4}$: $$\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}\left(\cos(x)\right)+i\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}(\sin(x))=\cos(x)+i\sin(x)=e^{ix}$$

*For $r\equiv 1\pmod{4}$: $$\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}(\cos(x))+i\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}(\sin(x))=-\sin(x)+i\cos(x)=ie^{ix}$$

*For $r\equiv 2 \pmod{4}$: $$\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}(\cos(x))+i\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}(\sin(x))=-\cos(x)-i\sin(x)=i^{2}e^{ix}$$

*For $r\equiv 3 \pmod{4}$: $$\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}(\cos(x))+i\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}(\sin(x)) = \sin(x)-i\cos(x)=i^{3}e^{ix}$$


And as $i^{r}\equiv i^{r\bmod{4}}$, we have:
$$\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}(\cos(x))+i\frac{\mathrm{d}^{r}}{\mathrm{d}x^{r}}(\sin(x))=i^{r}e^{ix}$$
