If you don't know how to get the derivative of inverse trigonometric functions, you can obtain the formula using implicit differentiation:
Let y = csc^-1 xy=csc−1x.
x = csc yx=cscy
d/dx x = d/dx csc yddxx=ddxcscy
1= dy/dx * -csc y cot y1=dydx⋅−cscycoty
(-1)/(csc y cot y)= dy/dx−1cscycoty=dydx
Use a fundamental identity to express all trig functions as the one you are differentiating, in this case: cot^2 theta + 1 = csc^2 thetacot2θ+1=csc2θ:
(-1)/(csc y sqrt(csc^2 y - 1) )= dy/dx−1cscy√csc2y−1=dydx
Use x = csc yx=cscy:
(-1)/(x sqrt(x - 1) )= dy/dx−1x√x−1=dydx
Therefore color(blue)(d/dx csc^(-1) x = (-1)/(x sqrt(x - 1) ))ddxcsc−1x=−1x√x−1
To simplify d/dx csc^-1(x^2 + 1)ddxcsc−1(x2+1) we can use the chain rule:
(f @ g)'(x) = f'(g(x)) * g'(x)
d/dx csc^-1(x^2 + 1)
= (-1)/((x^2 + 1) sqrt((x^2 + 1)- 1) ) d/dx (x^2 + 1)
= (-1)/((x^2 + 1) sqrt(x^2)) * (2x)
= (-2x)/((x^2 + 1) x)
= (-2)/(x^2 + 1)
