Answer
See the detailed answer below.
Work Step by Step
The on-axis potential of a charged disk is given by
$$V=\dfrac{Q}{2\pi \epsilon_0 R^2}\left[ \sqrt{R^2+z^2}-z\right]\tag 1$$
And we know that the electric field is given by
$$E=-\dfrac{dV}{dz}$$
Plug from (1),
$$E=-\dfrac{d }{dz}\left( \dfrac{Q}{2\pi \epsilon_0 R^2}\left[ \sqrt{R^2+z^2}-z\right]\right)$$
$$E=-\dfrac{Q}{2\pi \epsilon_0 R^2}\dfrac{d }{dz} \left[ (R^2+z^2)^\frac{1}{2}-z\right] $$
$$E=-\dfrac{Q}{2\pi \epsilon_0 R^2} \left[ \frac{1}{2} (2z) (R^2+z^2)^\frac{-1}{2}-1\right] $$
$$E=-\dfrac{Q}{2\pi \epsilon_0 R^2} \left[ \dfrac{z} {\sqrt{R^2+z^2}}-1\right] $$
$$\boxed{\vec E= \dfrac{Q}{2\pi \epsilon_0 R^2} \left[1- \dfrac{z} {\sqrt{R^2+z^2}} \right] \;\hat k}$$
