Derivative of arccsc(x)
-du/abs(u)sqrt(u^2-1)
Let y = arccsc(x), then csc(y) = x and y lies in the interval [0, π].
Now, we can use the chain rule to differentiate both sides with respect to x as follows:
d/dx [csc(y)] = d/dx [x]
Applying the chain rule, we get:
-csc(y)cot(y)dy/dx = 1
dy/dx = -cot(y)/csc(y)
To find the value of cot(y) and csc(y), we can use the identity:
csc^2(y) – 1 = cot^2(y)
Since csc(y) = x and y lies in the interval [0, π], we have:
sin(y) = 1/x and cos(y) = sqrt(1 – sin^2(y)) = sqrt(1 – 1/x^2)
Therefore, cot(y) = cos(y)/sin(y) = sqrt(1 – 1/x^2)/1/x = x/sqrt(x^2 – 1)
And csc(y) = x
Substituting these values into the expression above, we get:
dy/dx = -sqrt(x^2 – 1)/x^2
Therefore, the derivative of arccsc(x) is -sqrt(x^2 – 1)/x^2.
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