Question

How do I find the derivative of `y = csc^-1(x^2+1)`?

Answer 1

There is a pattern amongst the inverse trig derivatives:

`d/(dx)[arcsinu] = 1/sqrt(1-u^2)*((du)/(dx))`

`d/(dx)[arccosu] = -1/sqrt(1-u^2)*((du)/(dx))`

`d/(dx)[arc secu] = 1/(|u|sqrt(u^2-1))*((du)/(dx))`

`d/(dx)[arc cscu] = -1/(|u|sqrt(u^2-1))*((du)/(dx))`

`d/(dx)[arctanu] = 1/(1+u^2)*((du)/(dx))`

`d/(dx)[arc cotu] = -1/(1+u^2)*((du)/(dx))`

Notice though that Wolfram Alpha would say that the derivative of `arc cscu` is `-1/(sqrt(1-1/(u^2))*u^2`, but if you use its answer and manipulate it by calling the outer `u^2` equal to `|u||u| = |u|*sqrt(u^2)` and distributing it into the square root, it will be equivalent. We can use both, though.

`d/(dx)[arc csc(x^2+1)] = -1/(|x^2+1|sqrt((x^2+1)^2-1))*(2x)`

`= -(2x)/(|x^2+1|sqrt((x^2+1)^2-1))`

or...

`d/(dx)[arc csc(x^2+1)] = -1/((x^2+1)^2sqrt(1-1/(x^2+1)^2))*(2x)`

`= -(2x)/((x^2+1)^2sqrt(1-1/(x^2+1)^2))`