Derivative of ln(cscx - cotx)

We know the derivative of ln(x), which is ¹⁄ₓ.

[ln(x)]' = ¹⁄ₓ

We can find the derivative of ln(cscx - cotx) using chain rule.

Find ᵈʸ⁄d, if

y = ln(cscx - cotx)

Let t = cscx - cotx.

y = ln(t)

Now,

y = ln(t) ----> y is a function of t

t = cscx - cotx ----> is is a function of x

By chain rule, the derivative of y with respect to x :

Substitute y = ln(t) and t = cscx - cosx.

Substitute t = cscx - cosx.

Therefore,

[ln(cscx - cotx)]' = cscx

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