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
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

asked by amy

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