Deriving law of velocity in simple harmonic motion

Question

The traditional way of deriving velocity equation in SHM is:

But I wanted to derive it using pre-calculus and some basics. I couldn't pass this step:

Answer 1

The relationship relating a small increment in displacement to a small displacement in time is $\Delta x(t) = v(t) \Delta t + \mathcal{O} (\Delta t^2)$ . From this you can combine $a \Delta x = - \omega^2 x \Delta x $ with the previous expression to obtain $a v \Delta t = - \omega^2 xv \Delta t$.

Using the expressions for a and v we can write $av\Delta t= \frac{dv}{dt}v\Delta t= \frac{1}{2}\Delta v^2 $ etc then you obtain $\frac{1}{2}\Delta v^2 = -\frac{1}{2}\omega^2 \Delta x^2 + \mathcal{O}(\Delta^3)$. This expression only becomes exact in the limit as the increment goes to zero. It's not possible to derive the velocity expression without calculus since the natural language that newtonian mechanics is written in is calculus (note writing things like $\Delta x(t) = v(t) \Delta t + \mathcal{O} (\Delta t^2)$ is a taylor series expansion which comes from calculus so you are still using calculus in the second method) .