Higher order derivatives

Question

Let h(t)=tan(5t+8)

Then  is:

and  is:

Answer 1

Chain rule:

h' = d/dt (tan(5t + 8)) = sec²(5t + 8)*5 = 5*sec²(5t + 8)

Chain rule again:

h'' = d/dt 5*sec²(5t + 8) = 10*sec (5t + 8) * sec (5t + 8) * tan (5t + 8) * 5

= 50* sec² (5t + 8)*tan (5t + 8)

to find h'(4) and h"(4), simply plug in 4 to the equations. I assume that 4 is the angle in radians.