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