We have: #f(x) = csc(x) = frac(1)(sin(x))#
#Rightarrow f'(x) = frac(frac(d)(dx)(1) cdot sin(x) - frac(d)(dx)(sin(x)) cdot 1)(sin^(2)(x))#
#Rightarrow f'(x) = frac(0 cdot sin(x) - cos(x) cdot 1)(sin^(2)(x))#
#Rightarrow f'(x) = - frac(cos(x))(sin^(2)(x))#
#Rightarrow f'(x) = - frac(cos(x))(sin(x)) cdot frac(1)(sin(x))#
#Rightarrow f'(x) = - cot(x) cdot frac(1)(sin(x))#
#Rightarrow f'(x) = - frac(cot(x))(sin(x))#
Then, let's differentiate #f'(x)#:
#Rightarrow f''(x) = - (frac(frac(d)(dx)(cot(x)) cdot sin(x) - frac(d)(dx)(sin(x)) cdot cot(x))(sin^(2)(x)))#
#Rightarrow f''(x) = - (frac(- csc^(2)(x) cdot sin(x) - cos(x) cdot cot(x))(sin^(2)(x)))#
#Rightarrow f''(x) = - (frac(- (csc^(2)(x) sin(x) + cos(x) cot(x)))(sin^(2)(x)))#
#Rightarrow f''(x) = frac(csc^(2)(x) sin(x) + cos(x) cot(x))(sin^(2)(x))#
#Rightarrow f''(x) = frac(csc^(2)(x) sin(x))(sin^(2)(x)) + frac(cos(x) cot(x))(sin^(2)(x))#
#Rightarrow f''(x) = frac(csc^(2)(x))(sin(x)) + frac(cos(x))(sin(x)) cdot frac(cot(x))(sin(x))#
#Rightarrow f''(x) = csc^(2)(x) cdot csc(x) + cot(x) cdot cot(x) cdot csc(x)#
#Rightarrow f''(x) = csc^(3)(x) + cot^(2)(x) csc(x)#
One of the Pythagorean identities is #cos^(2)(x) + sin^(2)(x) = 1#.
If we divide through by #sin^(2)(x)#, we get:
#Rightarrow frac(cos^(2)(x))(sin^(2)(x)) + frac(sin^(2)(x))(sin^(2)(x)) = frac(1)(sin^(2)(x))#
#Rightarrow 1 + cot^(2)(x) = csc^(2)(x)#
#Rightarrow cot^(2)(x) = csc^(2)(x) - 1#
Let's apply this rearranged identity:
#Rightarrow f''(x) = csc^(3)(x) + (csc^(2)(x) - 1) csc(x)#
#Rightarrow f''(x) = csc^(3)(x) + csc^(3)(x) - csc(x)#
#therefore f''(x) = 2csc^(3)(x) - csc(x)#
