# 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