calculus
posted by amy .
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
dy/dx = dy/du*du/dx
dy/du = 1/u = 1/(sec x + tan x)
dy/dx = sec x tan x + sec^2 x
= sec x (sec x + tan x)
dy/dx = sec x
Respond to this Question
Similar Questions

Calculus  Integration
Hello! I really don't think I am understanding my calc hw. Please help me fix my errors. Thank you! 1. integral from 0 to pi/4 of (tanx^2)(secx^4)dx It says u = tan x to substitute So if I use u = tan x, then my du = secx^2 then I … 
math
1. (sinx/cscx)+(cosx/secx)=1 2. (1/sinxcosx)(cosx/sinx)=tanx 3. (1/1+cos s)=csc^2 scsc s cot s 4. (secx/secxtanx)=sec^2x+secxtanx 5. (cosx/secx1)(cosx/tan^2x)=cot^2x 
Calculus
find the derivative of f(x)=tanx4/secx I used the quotient rule and got (sec^2x4)(secx)(tanx4)(cosx)/sec^2x but I'm pretty sure that's wrong. Help. Thanks. 
Calculus
find the derivative y= csc^1 (secx) 0 less than or equal to x less than or equal to 2¤Ð what did i do wrong? 
Precalculus
Complete the following identity secx1/secx=? 
precal
1/tanxsecx+ 1/tanx+secx=2tanx so this is what I did: =tanx+secx+tanxsecx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)(1/cosx) =sinx/cosx+ sinx /cosx= 2tanxI but I know this can't be correct because what I did doesn't end as a negatvie:( … 
precal PLEASE HELP!
secx/secxtanx=sec^2x+ secx+tanx 
Math
I can't find the integral for (tanx)^(6)*(secx)^(2) I tried splitting up tanx into (tanx)^2*(tanx)^4 and let the latter equal (secx)^2  1. Please help, thanks! 
Gr 11 Functions
Simplify (1+tanx)^2 The answer is (1sinx)(1+sinx) Here's what I do: 1 + 2tanx + tan^2x When I simplify it becomes 1 + 2(sinx/cosx) + (1+secx)(1secx) What am I doing wrong? 
Math
Evaluate the integral of (secx)^2 * (tanx)^3 *dx I started out with letting u=secx and du=secx*tanx*dx , but then I am kind of stuck because I don't know how to factor out just secx*tanx?