We know the derivative of cscx, which is -cscxcotx.
(sinx)' = cscx
We can find the derivative of √cscx using chain rule.
If y = √cscx, find ᵈʸ⁄dₓ.
Let t = cscx.
Then, we have
y = √t
By chain rule,
Substitute t = cscx.
Therefore,
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