Answer
$\frac{d}{dx}[cos x] = \lim\limits_{Δx \to 0} \frac{cos(x + Δx) - cosx}{Δx} = \lim\limits_{Δx \to 0} \frac{cosx*cosΔx - sinx*sinΔx-cosx}{Δx} = \lim\limits_{Δx \to 0} \frac{(cosx*cosΔx -cosx)- sinx*sinΔx}{Δx} = \lim\limits_{Δx \to 0} [\frac{-cosx (1-cosΔx)}{Δx} - \frac{sinx*sinΔx}{Δx} ] = -cosx (\lim\limits_{Δx \to 0} \frac{1-cosΔx}{Δx}) - sinx (\lim\limits_{Δx \to 0} \frac{sinΔx}{Δx}) = [-cosx (0) - sinx (1)] = -sinx$
Work Step by Step
To prove that $\frac{d}{dx}[cosx ]= -sinx$ we can do the following steps: $\frac{d}{dx}[cos x] = \lim\limits_{Δx \to 0} \frac{cos(x + Δx) - cosx}{Δx} = \lim\limits_{Δx \to 0} \frac{cosx*cosΔx - sinx*sinΔx-cosx}{Δx} = \lim\limits_{Δx \to 0} \frac{(cosx*cosΔx -cosx)- sinx*sinΔx}{Δx} = \lim\limits_{Δx \to 0} [\frac{-cosx (1-cosΔx)}{Δx} - \frac{sinx*sinΔx}{Δx} ] = -cosx (\lim\limits_{Δx \to 0} \frac{1-cosΔx}{Δx}) - sinx (\lim\limits_{Δx \to 0} \frac{sinΔx}{Δx}) = [-cosx (0) - sinx (1)] = -sinx$ Then, is proved that $\frac{d}{dx}[cosx ]= -sinx$
