For the case of the moving light-clock, the path traveled by the light (L) is related to the distance the clock moves (d) and the distance between the two mirrors (l) by the Pythagorean theorem:
\begin{equation} L^2 = l^2 + (\frac{d}{2})^2 \end{equation}
We have written d/2 because the right triangle is only half of the light’s path.
We can write these distances in terms of v, the time intervals, and c, to derive an expression for Δt' in terms of Δt, v, and c.
Writing d in terms of v and Δt':
\begin{equation} L^2 = l^2 + v^2(\frac{\Delta t’}{2})^2 \end{equation}
Traveling the distance L takes only half the time between ticks because the light has to traverse it once on the way up and then again on the way down to make one tick. That is what the factor of two represents in the expression for d: we have only traveled half of the full time interval needed for a clock tick by traveling up to the top mirror. Since we assume that the speed of light is constant (according to the second principle of relativity), we can write a relation between L, c, and Δt' like the one we wrote for l, c, and Δt:
\begin{equation} \Delta t’ = \frac{2L}{c} \end{equation}
This looks similar to our expression for l (as measured when the clock is at rest), but it is for the case of the moving clock. Solving these for L and l and substituting them into the equation for L2, we get:
\begin{equation} (\frac{c\Delta t’}{2})^2 = (\frac{c\Delta t}{2})^2 + (\frac{v\Delta t’}{2})^2 \end{equation}
Now rearranging by collecting the common time terms we get:
\begin{equation} c^2 \Delta t’^2 (1- \frac{v^2}{c^2}) = c^2 \Delta t^2 \end{equation}
And finally, canceling the common term of c2 and taking the square root:
\begin{equation} \Delta t’ = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}} \end{equation}