f(x)=secx/1+secx?

Question

find the derivative. (differenctiate)

Answer 1

1st: using the trigonometric identity sec x = 1/cos x, we simplify
f(x) as follows:
f(x) = sec x/(1 + sec x) = (1/cos x)/(1 + (1/cos x))
= 1/(1 + cos x)
2nd: differentiate the resulting expression for f(x) w.r.t. x:
f'(x) = (-1)(1 + cos x)^(-2)(-sin x)
f'(x) = (sin x)/(1 + cos x)^2

Answer 2

This differentiation is made both shorter and easier by first simplifying f(x):

f(x) = (1/cos x)/(1 + 1/cos x)

(Im presuming you meant to put braces around "1 + sec x"; it looks much better and much more suggestive for the next step when written out on the page with "____"symbols for division rather than "/".)

It simplifies to f(x) = 1/(1 + cos x) or (1 + cos x)^(-1)

This is now trivial to differentiate. We have:

df/dx = sin x /(1 + cos x)^2; that's it!

Note that there was no need to mess around with sec and tan x's, the combining of separate terms, or anything else as in the first responder's otherwise correct but more circuitous method. In addition, by using the squared form in the denominator above, the solution is written in its most compact form.

Live long and prosper.

Answer 3

OK Uber... braket mismatch.. I try again

sec/(1+sec)=1/(1+cos)

f'(x)=sin/(1+cos)²

Answer 4

capital F as opposed to lower case f is usually used for a function that has been integrated already....

so usually,

F'(x) = f(x) meaning you just get the function on the right hand side.

Answer 5

[(1+secx)(d/dx secx)-secx(d/dx(1+secx)]/(1=secx)^2
= [(1+secx)(secxtanx)-secx(secxtanx)]/(1+secx)^2
= [secxtanx+sec^2xtanx-sec^2xsecxtanx]/(1+secx)^2
=secxtanx/(1+secx)^2
=sinx/(1+cosx)^2

Answer 6

sec(x)*tan(x)/(1+sec(x))-sec(x)^2*tan(x)/(1+sec(x))^2
simplified it equals sin(x)/(cos(x)^2+2*cos(x)+1)
Do not listen to champ, sec = 1/cos but sec/1+sec = (1/cos)/(1+(1/cos)) not 1+cos/cos