Derivative of cosine

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}x} \cos(x) = -\sin(x),$$ where $\cos$ denotes the cosine and $\sin$ denotes the sine.

Proof

From the definition of cosine, $$\cos(z) = \dfrac{e^{iz}+e^{-iz}}{2},$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, the fact that $\dfrac{1}{i}=-i$, and the definition of the sine function, $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) &= \dfrac{1}{2} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] + \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\ &= \dfrac{1}{2} \left[ ie^{iz} - ie^{-iz} \right] \\ &= -\dfrac{e^{iz}-e^{-iz}}{2i} \\ &= -\sin(z), \end{array}$$ as was to be shown. █