Question

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

Answer 1

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`