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I have a derivative question?
Been working on this for a while and can't figure it out.... need the derivative of (lnx)^(2secx)
2 Answers
- grunfeldLv 71 decade ago
y = ( ln x )^(2 sec x )
ln y = (2 sec x ) ln ( ln x )
(1 / y ) dy / dx = (2 sec x ) [ 1 / ( x ln x ) ] + ( 2 sec x tan x ) ln ( ln x )
dy / dx = ( 2 sec x ) ( ln x )^(2 sec x ) / ( x ln x ) + (2 sec x tan x ) ln ( ln x ) ( ln x )^(2 sec x )
Source(s): my brain - 1 decade ago
Let's implicitly derive
y = ln(x)^(2sec(x))
ln(y) = ln(ln(x)^(2sec(x))
ln(y) = 2 * sec(x) * ln(ln(x))
dy / y = 2 * (sec(x) * (1/x) * (1/ln(x)) + ln(ln(x)) * sec(x) * tan(x)) * dx
dy / dx = 2y * sec(x) * (ln(ln(x)) * tan(x) + 1 / (xln(x)))
dy / dx = 2 * ln(x)^(2sec(x)) * sec(x) * (ln(ln(x)) * tan(x) + 1 / (xln(x)))
