# Math

posted by .

f=8ln(sec(x)+tan(x))
f'=

• Math -

f'=8sec(x)

## Similar Questions

1. ### Integration

Intergrate ¡ì sec^3(x) dx could anybody please check this answer. are the steps correct?
2. ### Calculus

could anybody please explain how sec x tan x - ¡ì sec x tan^2(x) dx = sec x tan x + ¡ì sec x dx - ¡ì sec^3(x) dx What I don't understand about your question is what is ¡ì ?
3. ### more trig.... how fun!!!!

if you can't help me with my first question hopw you can help me with this one. sec(-x)/csc(-x)=tan(x) thanx to anyone who can help From the definition of the sec and csc functions, and the tan function, sec(-x)/csc(-x) = sin(-x)/cos(-x) …
4. ### calculus

find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it. Use the chain rule. Let y(u) = ln u u(x) = sec x + tan x …
5. ### calculus

Use integration by parts to evaluate the integral of x*sec^2(3x). My answer is ([x*tan(3x)]/3)-[ln(sec(3x))/9] but it's incorrect. u=x dv=sec^2(3x)dx du=dx v=(1/3)tan(3x) [xtan(3x)]/3 - integral of(1/3)tan(3x)dx - (1/3)[ln(sec(3x))/3] …
6. ### Pre-Cal

Perform the addition or subtraction. tanx - sec^2x/tanx tan^2(x)/tan(x) - sec^2x/tanx = tan^2x sec^2x / tanx then I use the identity 1+tan^2u=sec^2u I do not know what to do at this point.