if y=√2f(t)+1 f(0)=0 and f'(0)=4 find dy/dt...t=0

Question

if

y=√2f(t)+1

f(0)=0 and f'(0)=4

find dy/dt...t=0

Answer 1

I'm going to assume that everything on the right-hand side is meant to be under the radical: y = √(2f(t)+1)

We need to use chain rule here, when we are not given a specific function rule (equation) for f(t). But that won't matter since we are given the values of f(0) and f^'(0), which are all we'll need.

Chain rule says take the derivative of the "inside" function, in our case 2f(t)+1, then multiply that derivative by the derivative of the "outside" function, treating whats inside as if it were just x or something:

dy/dt = 2f^'(t)·(1/2)·(1/√(2f(t)+1)) = f^'(t) / √(2f(t)+1)

dy/dt_t=0 = f^'(0) / √(2f(0)+1) = 4 / √1 = 4