what is the derivitive of 1+tan^2x?

Question

I need the derivitive of this using the chain rule, but the answer has to be in terms of tangent, not secant, which is the derivitive of tan

i am working with a maclaurin series question and the prof gave us the following table of derivatives for y=tan(x)
k f^k(x)
0 tan(x)
1 1 + tan^2(x)
2 2(tan^3(x) +tan x)
3 6tan^4x + 8 tan^2 +2
4 24tan^5x + 40tan^3x + 10tanx
5 120tanx^6+240tan^4 +136tan^2x

I don't understand how he getting each derivative

Answer 1

y = 1 + (tan x) ²
dy/dx = 2 tan x sec ² x
dy/dx = 2 tan x (tan ² x + 1)
dy/dx = 2 tan ³ x + 2 tan x

Answer 2

y =1+tan^2x
dy/dx = 2(tanx) . sec^2x
= 2tanx / cos^2x
=2sinx / cos^3x

Another way
dy/dx = 2(tanx) . sec^2x
= 2tanx {tan^2x +1}
=2 tan^3 x +2tanx


Additional Details

All he is giving you here are the successive derivatives of tanx, using d(tanx)/dx = sec ^2 x as he goes and using the Pythagorean Identity that
sec^2 x = 1 + tan^2 x
to replace sec^2 x whenever it appears.


y=tanx
y' =sec^2 = 1 +tan^2 x
y'' = 2 tan ^3x +2tanx from above

y''' = 6(tanx)^2.sec^2 x +2 (sec^2x)
=6 tan^2x ( 1+tan^2 x) + 2(1+tan ^2x)
=8tan^2 x + 6tan ^4 x +2


y''''=16 tanx. sec^2x +24tan^3 x . sec^2 x
= 16tanx (1+tan^2 x) +24 tan^3 x( 1 +tan^2 x)
=16tanx +40 tan^3 x + 24 tan^5 x


and so on

Answer 3

the derivitive of 1+tan^2x

since the derivitive of any integer is 0, we can ignore the one

since tan^2x is basically (tanx)^2
we first find the derivitive of (function)^2
which is basically 2(functionx)^1

so tan^2x = 2(tanx)
we're not done yet though, we have to find the derivitive of the function inside.
(tanx)' = sec^2x
so

(1+tan^2x)' = 2(tanx)*sec^2x

Answer 4

y=1 +tan^2 x
dy=2tanx*sec^2 x dx
dy/dx=2tan x *sec^2 x

Answer 5

derivative not "derivitive"!!!

1 + tan^2(x) = sec^2(x)

just write tan(x)=sin(x)/cos(x)
sec(x)=1/cos(x)
and use sin^2(x)+cos^2(x)=1