We know the derivative of lnx, which is ¹⁄ₓ. And also, the derivative cscx is -cscxcotx.
(lnx)' = ¹⁄ₓ
(-cscx)' = -cscxcotx
We can find the derivative of ln(cscx) using chain rule.
If y = ln(cscx), find ᵈʸ⁄dₓ.
Let t = cscx.
Then, we have
y = lnt
By chain rule,
Substitute y = lnt and t = cscx.
Substitute t = cscx.
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