We know the derivative of ln(x), which is ¹⁄ₓ.
[ln(x)]' = ¹⁄ₓ
We can find the derivative of ln(secx + tanx) using chain rule.
Find ᵈʸ⁄dₓ, if
y = ln(secx + tanx)
Let t = secx + tanx.
y = ln(t)
y = ln(t) ----> y is a function of t
t = secx + tanx ----> t is is a function of x
By chain rule, the derivative of y with respect to x :
Substitute y = ln(t) and t = secx + tanx.
Substitute t = secx + tanx.
[ln(secx + tanx)]' = secx
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