1. I am supposed to find the derivative of (sqrt) 3x using the (lim h ->0) f'(x) = (f(x+h) - f(x))/h equation and here is what i got:
((sqrt) 3(x+h) - (sqrt) 3x) / h
then i got ((sqrt) 3x + 3h) - (sqrt) 3x) / h
I then used the conjugate of this and got:
(3x + 3h - 3x) / h((sqrt) 3x + 3h + (sqrt) 3x)
the 3x in the numerator cancels and the h in 3h is factored out. so i then get:
3 / ((sqrt) 3x + 3h) + (sqrt) 3x.
when h -> 0, i got
3 / 2(sqrt) 3x.
I know I am making a dumb mistake somewhere and know that the answer is suppose to be: (3/2) x^(-1/2)
2. This is a problem involving an airplane taking off...Suppose that the distance an aircraft travels along a runway before takeoff is given by D = (10/9)t^2 where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km/h. How long will it take to become airborne, and what distance will it travel in that time?
The problem I have with this is I do not know where to start nor end.
3. A particle moves on a coordinate line such that it has a position
s = 2t^3 - 4t^2 + 7 on the interval 0 =< t =< 3, with s measured in meters and t measured in seconds. Find the particle's velocity and acceleration at the endpoints of the interval. This is what i got.
s'(t) = 6t^2 - 8t
s''(t) = 12t - 8
v(0) = 6(0)^2 - 8(0) = 0 m/sec.
v(3) = 54 - 24 = 30 m/sec
A(0) = 12(0) - 8 = -8 m/sec
A(3) = 12(3) - 8 = 28 m/sec
I just need to know if this is right.
4. I am suppose to find the derivative of 2t^(3/2) + 3e^2
e is the natural log.
I know the first part is 3t^(1/2), but i do not know about the second part. The answer in the book said just 3t^(1/2). But I thought that I am suppose to get something for an answer for the second part and this is where i got confused.
Thank you in advance.