To evaluate inverse trigonometric functions, a valuable technique is implicit differentiation:
d/dxf(y) = d/dyf(y)dy/dxddxf(y)=ddyf(y)dydx
In this case, we also will use the chain rule:
d/dxf(g(x)) = f'(g(x))g'(x)
as well as the following derivatives:
d/dxcsc(x) = -csc(x)cot(x)
d/dxx^n = nx^(n-1)
Now, let y = "arccsc"(sqrt(x+1))
=> csc(y) = csc("arccsc"(sqrt(x+1))) = sqrt(x+1)
=> d/dxcsc(y) = d/dxsqrt(x+1)
Through implicit differentiation:
d/dxcsc(y) = d/dycsc(y)dy/dx = -csc(y)cot(y)dy/dx
And through the chain rule
d/dx(x+1)^(1/2) = 1/2(x+1)^(-1/2)(d/dx(x+1)) = 1/(2sqrt(x+1))
So
-csc(y)cot(y)dy/dx = 1/(2sqrt(x+1))
=> dy/dx = -1/(2sqrt(x+1)*csc(y)cot(y))
But we want our final answer entirely in terms of x. To fix this, we remember that csc(y) = sqrt(x+1) and draw a corresponding right triangle:
From this we can see that cot(y) = sqrt(x)
Thus we have
-1/(2sqrt(x+1)*csc(y)cot(y)) = -1/(2sqrt(x+1)*sqrt(x+1)*sqrt(x))
From this, we can get our final answer
d/dx "arccsc"(sqrt(x+1)) = -1/(2(x+1)sqrt(x))
