To evaluate inverse trigonometric functions, a valuable technique is implicit differentiation:

#d/dxf(y) = d/dyf(y)dy/dx#

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))#