Question

derivative of (2t^3/2) + 25 using first principle

Answer 1

more?
(f(t+h)-f(t))/h = (2(t+h)^(3/2) + 25 - 2t^(3/2) - 25)/h
= 2((t+h)^(3/2) - t^(3/2))/h
now multiply top and bottom by the radical conjugate
= 2((t+h)^(3/2) - t^(3/2))((t+h)^(3/2) + t^(3/2)) / h((t+h)^(3/2) + t^(3/2))
= 2((t+h)^3 - t^3) / h((t+h)^(3/2) + t^(3/2))
= 2(3ht^2 + 3h^2t + h^3) / h((t+h)^(3/2) + t^(3/2))
= 2(3t^2+3ht+h^2) / ((t+h)^(3/2) + t^(3/2))
Now take the limit as h -> 0 and you have
2(3t^2) / 2t^(3/2)
= 3t^(1/2)