How do you find the derivative of Inverse trig function #f(x) = arcsin (9x) + arccos (9x)#?

1 Answer
Jun 21, 2015

Here'/ the way I do this is:
- I'll let some #" "theta=arcsin(9x)" "# and some #" "alpha=arccos(9x)#

  • So i get, #" "sintheta=9x" "# and #" "cosalpha=9x#

  • I differentiate both implicitly like this:
    #=>(costheta)(d(theta))/(dx)=9" "=>(d(theta))/(dx)=9/(costheta)=9/(sqrt(1-sin^2theta))=9/(sqrt(1-(9x)^2)#

-- Next, I differentiate #cosalpha=9x#
#=>(-sinalpha)*(d(alpha))/(dx)=9" "=>(d(alpha))/(dx)=-9/(sin(alpha))=-9/(sqrt(1-cosalpha))=-9/sqrt(1-(9x)^2)#

  • Overall, #" "f(x)=theta+alpha#

  • So, #f^('')(x)=(d(theta))/(dx)+(d(alpha))/(dx)=9/sqrt(1-(9x)^2)-9/sqrt(1-(9x)^2)=0#