I can't get this:
y = ln(1 + tan x)
(i) Show that d2y/dx2 = -1 when x = 0.
Thanks if you could help, I'm preparing for my promos next week!
Step 1 use the chain rule:
dy/dx=[1/(1+tan(x))](sec(x))^2
Step 2 use product rule (you could use the quotient ruleif you prefer)
and chain rule:
d^2y/dx^2=[1/(1+tan(x))^2](-1) [sec(x)]^4 + [1/(1+tan(x))].2.[sec(x)]^2 tan(x)
when x=0 tan(x)=0, and sec(x)=1, result follows.(i) Show that d2y/dx2 = -1 when x = 0.
RonL
Hello, margaritas!
It's just two applications of the Quotient Rule . . .
. . Exactly where is your difficulty?
y = ln(1 + tan x)
(i) Show that d²y/dx² = -1 when x = 0
dy . . . . . .sec²x
--- . = . ------------
dx . . . . 1 + tan x
d²y . . . (1 + tan x)·2·sec x(sec x·tan x) - sec²x·sec²x
---- .= .------------------------------------------------------
dx² . . . . . . . . . . . . . (1 + tan x)²
. . . . . . 2·sec²x·tan x(1 + tan x) - sec²x·sec²x
. . . = . ----------------------------------------------
. . . . . . . . . . . . . (1 + tan x)²
When x = 0, we have:
. . . d²y . . . . 2·sec²0·tan 0(1 + tan 0) - sec²0·sec²0
. . . ---- . = . ----------------------------------------------- . = . -1
. . . dx² . . . . . . . . . . . (1 + tan 0)²