DERIVATIVE OF NATURAL LOG OF CSCX

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.

Therefore,

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