First note that f'(x) denotes the first derivative of a function while f(x) denotes a function whose first derivative is to be taken. One would then state the problem as "Find the derivative of f(x) = arccsc [8x]".
For y = arccosecant 8x, set down csc y = 8x.
Then d(csc y)/dx = d(8x)/dx.
Rewrite this last as d(1/sin y)/dx = d(8x)/dx.
The Differentiation Quotient Rule gives d(1/sin y)/dx as
[(sin y)(0)−(1)(cos y)/sin2y](dy/dx) or [-cos y/sin2y]dy/dx
with d(8x)/dx going to 8 on the right side.
Now rewrite [-cos y/sin2y]dy/dx = 8 as
(-1/sin y)(cos y/sin y)dy/dx = 8 or
(-csc y)(cot y)dy/dx = 8.
Then dy/dx = -8 ÷ (csc y)(cot y).
Search for substitutions while noting that csc2y = 1 + cot2y
which gives cot y = ±√{csc2y − 1}.
Now transform dy/dx = -8 ÷ (csc y)(cot y) to
dy/dx = -8 ÷ [8x • ±√{(8x)2 − 1}].
Simplify this last to dy/dx = ±1 ÷ [x • √{64x2 − 1}].
A graph of arccosecant 8x shows the slopes of lines tangent to
the graph curves (also known as the first derivatives)
to be negative.
The derivative then sought is dy/dx = -1 ÷ [x • √{64x2 − 1}] for |x| ≥ 1/8.
