find dy/dx...? Calculus 1 help...?

Question

can u help me
my teacher does one example and thats it...
y=sec^2*x - tan^2*x


bonus
find a formula for the nth derivative.
y=1/(Square root of x)

Answer 1

First one: simplify first!
There's an identity that says sec^2 x = tan^2 x + 1. So y = 1 and dy/dx = 0.

Otherwise, if you differentiate directly, you'll get
dy/dx = 2 sec x (sec x tan x) - 2 tan x (sec^2 x) = 0
if you know d/dx sec x = sec x tan x and d/dx tan x = sec^2 x.

For the bonus question:
y = x^(-1/2)
y' = (-1/2) x^(-3/2)
y'' = (-1/2) (-3/2) x^(-5/2)
y''' = (-1/2) (-3/2) (-5/2) x^(-7/2)
...
y(n) = (-1/2) ... (-(2n-1)/2) x^(-(2n+1)/2)
= (-1/2)^n (1 . 3 . ... (2n-1)) x^(-(2n+1)/2)
= (-1/2)^n (2n-1)!! x^(-n-1/2)
where (2n-1)!! is the "double factorial" of (2n-1), defined essentially as
x!! = x (x-2) (x-4) ... 5 . 3 . 1 for x odd,
x!! = x (x-2) (x-4) ... 6 . 4 . 2 for x even.

Answer 2

First, carry the (a million/3) out in front, subtract one from the exponent to get (-2/3) and go away the interior on my own. Then distinctive via the spinoff of the interior, which fits to be an straightforward quotient rule. stable success!