Question

How do you find the derivative of Inverse trig function `y = tan(x + sec x)`?

Answer 1

`y^' = sec^2(x+secx) * (1 + secxtanx)`

Explanation:

All you really need to use in order to differentiate this function is the chain rule for `tanu`, witrh `u = x + secx`, provided of course that you what the derivatives of `tanx` and `secx` are.

`color(blue)(d/dx(tanx) = sec^2x)" "` and `" "color(blue)(d/dx(secx) = secx * tanx)`

So, the derivative of `y` will be

`d/dx(y) = d/(du)(tanu) * d/dx(u)`

`y^' = sec^2u * d/dx(x + secx)`

`y^' = sec^2u * (1 + secxtanx)`

`y^' = color(green)(sec^2(x+secx) * (1 + secxtanx))`