## Thomas' Calculus 13th Edition

$y'=-3x^{1/2}(x+1)^2csc(x+1)^3cot(x+1)^3+\frac{1}{2}x^{-1/2}csc(x+1)^3$
Rewrite the equation: $y=x^{1/2}csc(x+1)^3$ Take the derivative of the equation using Power Rule, Product Rule, Trigonometric derivatives, and Chain Rule: $y'=x^{1/2}\times-csc(x+1)^3cot(x+1)^3\times3(x+1)^2\times1+\frac{1}{2}x^{-1/2}csc(x+1)^3$ $=-3x^{1/2}(x+1)^2csc(x+1)^3cot(x+1)^3+\frac{1}{2}x^{-1/2}csc(x+1)^3$