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

Question

3106 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-21T23:44:49

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

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