what is the derivative of sec^2(x)/tan(x)?

Answer 1

let y = sec²x/tanx
Now sec²x = tan²x + 1
So y = (tan²x + 1)/tanx
= tanx + cotx
So dy/dx = sec²x - cosec²x

Answer 2

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

now it can be differentiated as

- cosec^2 x + sec ^2 x

Answer 3

Look at it this way;
sec = 1/cos
tan = sin/cos
So sec2 = cos/sin = 1/tan = cot.
That should be on one of your tables.

Answer 4

We have:
sec² x/tan x
1/cos² x * cos x/sin x
1/(sin x cos x)
(sin x cos x)^(-1)

Using the chain rule:

dy/dx = -(sin x cos x)^(-2) * (cos² x - sin² x)
sin²x/(sin² x cos² x) - cos² x/(cos² x sin² x)
sec² x - csc² x

Answer 5

sec^2(x)=1/cos^2(x)------------(i)
and
tan(x)=sin(x)/cos(x)------------(ii)
from (i) and (ii)
sec^2(x)/tan(x)={1/cos^2(x)}/{sin(x)/cos(x)}

={1/cos^2(x)}X{cos(x)/sin(x)}

=cos(x)/cos^2(x)sin(x)

=1/{sin(x)cos(x)}

[sin(2x)=2sin(x)cos(x) therefor {sin(x)cos(x)} =sin(2x)/2

because of that
y= sec^2(x)/tan(x)=1/[sin(2x)/2]

y =2/sin(2x)

y =2cosec(2x)
then we can diffrenshad it
dy/dx=-{2cosec(2x)cot(2x)}X{d(2x)/dx}

=-4cosec(2x)cot(2x)

=-4{1/sin(2x)}{cos(2x)/sin(2x)}

=-4cos(2x)/sin^2(2x)

now it,s finished.like this you must make that large fountain to a small fountain.then it will make eazy to find dy/dx.