DERIVATIVE OF SQUARE ROOT OF CSCX

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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