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

1 Answer
Stefan V.
Aug 26, 2015

#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))#

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