Calculus
z′=cos(t+z)−1z^{\prime}=\cos\left(t+z\right)-1
z′=sec(t+z)+1z^{\prime}=\sec\left(t+z\right)+1
z′=sec(t+z)−1z^{\prime}=\sec\left(t+z\right)-1
z′=cos(t+z)+1z^{\prime}=\cos\left(t+z\right)+1