Related rates problem... ship A and B leave the same port?

Let t = number of hours elapsed since noon
Since A leaves at noon and its speed is 20, its distance traveled is 20t
Since B leaves at 1pm, it has been traveling for t-1 hours, and its speed is 15, so its distance traveled is 15(t-1) = 15t - 15
Since there is a 90 degree angle between the trajectories of the two ships (west and south), they form the two legs of a right-angled triangle. The direct distance between the ships is the hypotenuse. So we use Pythagoras:
d^2 = (20t)^2 + (15t - 15)^2
d^2 = 400t^2 + 225t^2 - 450t + 225
d^2 = 625t^2 - 450t + 225
d = sqrt(625t^2 - 450t + 225), note that a distance (d) can't be negative
Let u = 625t^2 - 450t + 225, so du/dt = 1250t - 450
So d = sqrt(u) = u^0.5, so dd/du = 0.5u^(-0.5) = 1 / (2 * sqrt(625t^2 - 450t + 225))
By the Chain Rule:
dd/dt = dd/du * du/dt
dd/dt = (1 / (2 * sqrt(625t^2 - 450t + 225))) * (1250t - 450)
dd/dt = (625t - 225) / sqrt(625t^2 - 450t + 225)
At 2pm, t = 2, so we have:
dd/dt = (625*2 - 225) / sqrt(625*2^2 - 450*2 + 225)
dd/dt = (1250 - 225) / sqrt(625*4 - 900 + 225)
dd/dt = 1025 / sqrt(2500 - 900 + 225)
dd/dt = 1025 / sqrt(1825)
dd/dt = 205 / sqrt(73)
dd/dt =~ 23.993435175206765608406683655879

A leaves at noon. At 2:00 pm, it would have traveled 2*20 = 40 knots. B leaves at 1:00pm. At 2:00 pm, it would ave traveled 15 knots.
x^2+y^2=z^2
40^2+15^2 = z^2
z = sqrt(40^2 +15^2) = 42.72

x^2+y^2 = z^2
2x dx/dt + 2y dy/dt = 2z dz/dt
x dx/dt + y dy/dt = z dz/dt
dz/dt = ( x dx/dt + y dy/dt) / z

dz/dt = ((40)(20) + (15)(15)) /42.72 = 23.99 knots/hr
https://gyazo.com/3d60156f194d433d5d26df8f835505ff