# calculus

posted by .

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

## Similar Questions

1. ### 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 …
2. ### math

1. (sinx/cscx)+(cosx/secx)=1 2. (1/sinxcosx)-(cosx/sinx)=tanx 3. (1/1+cos s)=csc^2 s-csc s cot s 4. (secx/secx-tanx)=sec^2x+secxtanx 5. (cosx/secx-1)-(cosx/tan^2x)=cot^2x
3. ### Calculus

find the derivative of f(x)=tanx-4/secx I used the quotient rule and got (sec^2x-4)(secx)-(tanx-4)(cosx)/sec^2x but I'm pretty sure that's wrong. Help. Thanks.
4. ### 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?
5. ### Calculus

I think I have the correct answers for the following problems. For anyone who has the time I would appreciate it if you could tell me if I am correct/incorrect. Thank you. 1). Differentiate ã(x)sin x = (2xcosx+sinx)/2ã(x) 2). …
6. ### Precalculus

Complete the following identity secx-1/secx=?
7. ### precal

1/tanx-secx+ 1/tanx+secx=-2tanx so this is what I did: =tanx+secx+tanx-secx =(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:( …