How do u prove the derivative of arc csc(x) and arc cotangent(x)?

Question

I would really appreciate it if you could at least tell me the identity in terms of square root of -csc(x)cotan(x) and -csc(x)^2(acotan(x))

Answer 1

Let arc csc x = arcsin (1/x)
now use formula for arcsin with chain rule

Let arc cot x = arctan (1/x)
now use formula for arctan with chain rule
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You can also use implicit differentiation:
y = arc csc x is equivalent to x = csc y
Now do derivaitve of this implicitly:
x = csc y
1 = - cscy coty (dy/dx)
dy/dx = -1 / (cscycoty)

Now to get in terms of x, draw triangle from original ratio
csc y = hypotenuse/opposite = x/1
adjacent found by Pyth theorem
three sides of triangle with angle y :
opp = 1
hyp = x
adj = sqrt(x^2 - 1)

so csc y = x / 1
and cot y = sqrt(x^2 - 1) / 1

From above
dy/dx = -1 / (csscycoty)
dy/dx = -1 / [ x sqrt(x^2-1) ]

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Answer 2

Consider that (f^-1)'(x) = 1/f'((f^-1)(x) when f'(x) is not 0.