Given x=2(t^2)+1 and y=3(t^3)+2, find the second 
derivitive [(d^2)y]/[d(x^2)] at t=1.  Please show work.

I believe the answer is 4 1/2, but I'm far too many years removed from
Calculus study to have faith in that...

I SHOULD know this...but alas...

I think it's a parametric equation, being that it's a function of two
equations....dats about it...

As I recall, the derivative of a(x^b) is ba(x^b-1) (where a is a
constant, b is an integer, and x is unknown) and the derivative of a
constant is zero (because the x would be x to the zeroth power, and
the next derivative would require multiplying by zero, resulting in
zero)

So first we expand out the equations above:
   x  = 2t^2 + 2
   y  = 3t^3 + 6

The first derivative would be:
   x' = 4t^1  (or just 4t)
   y' = 9t^2

And the second derivative would be:
   x''= 4
   y''= 18t^1 (or just 18t)

At t=1, x=4 and y=18, so the coordinates would be (4,18)

Plotting for any value t, I believe you'd get a straight vertical line at x=4.

There are lots of ways to do this problem.  Here's one:

x = 2t^2 + 1  -->  dx/dt = 4t  -->  dt/dx = 1/(4t)

y = 3t^3 + 2  -->  dy/dt = 9t^2

dy/dx = dy/dt dt/dx = 9t^2 / (4t) = (9/4)t

d^2y/(dx)^2 = (9/4) dt/dx = 9/(4*4t) = 9/(16t) = (9/16)t^(-1)

So at t=1, the required second derivitive is equal to 9/16.

There are many ways to solve this problem.
Here is one

x=2(t^2)+1 and y=3(t^3)+2
When t=1 we have x=3 and y=5.

x=2t^2+1 is equivalent to t^2=(x-1)/2
When x>1 we get t=((x-1)/2)^(1/2)
Consequently y=3((x-1)/2)^(3/2)+2

Let us now differentiate this function twice and compute the derivative at x=3.

dy/dx=(3/2)*3*(1/2)*((x-1)/2)^(1/2)=9/4*((x-1)/2)^(1/2)
d^2y/dx^2=9/4*(1/2)*(1/2)*((x-1)/2)^(-1/2)=9/16*((x-1)/2)^(-1/2)

at x=3 the derivative is 9/16*((3-1)/2)^(-1/2)=9/16