Calculus help!!?

Question

Find the derivative:

1) g(t)= sq. rt.(1/t^2-2)

2) h(x)= sin 2x cos 2x

And did I do these two right?
Find an equation of the tangent line to the graph of f at the given point:
3) y=2tan^3, (pi/4,2)
y^1= 6(tan^2x)(sec^2x)
= 6(tan(pi/4)^2)(sec(pi/4)^2)
= 6(1)(2)
=12
Find the 2nd derivative of the function:
4) f(x)= 1/x-2
f^I(x)= -1x^-2
f^II(x)= 2x^-3

Answer 1

1) g(t)= sq. rt.(1/t^2-2) = (t^2-2)^(-1/2)
g'(t) = -(1/2)(2t)(t^2-2)^(-3/2) = -t(t^2-2)^(-3/2)

2) h(x) = (1/2)sin 4x
h'(x) = 2cos 4x

3) You did right to get y' at (pi/4,2).

4) Your answer would be right if f(x) = 1/x

Answer 2

1) g(t)= sq. rt.(1/t^2-2)
g' = [1/(2sqrt(1/t^2-2)](-2/t^3)


2) h(x)= sin 2x cos 2x
h' = sin(2x) (-2sin(2x) +2cos(2x)cos(2x)
h' = 2 (cos^2 (2x) - sin^2 (2x)
h' = 2cos^(4x)

3) You did not finish
y= 12x +b
2= 12(pi/4) +b
b = 2- 3pi
y = 12x +2-3pi is equation of tangent

4) is fine