Question

use logarithmic differentiation to find the derivative of the function: y=(cos 2x)^x.
For an answer i got:
(cos2x)^x[(x)(-2tan(x))+ln(cos2x)(1/x)] but it was wrong.

Answer 1

take log on both sides,

1/y* dy/dx = d(x * log cos(2x))/dx

dy/dx = y * d(x * log cos(2x))/dx

dy/dx = (cos 2x)^x * d(x * log cos(2x))/dx

Answer 2

y=(cos 2x)^x.
Take ln on both sides
lny = ln (cos 2x)^x
ln y = x ln cos 2x
diff. both side w.r.t. x
(1/y)(y') = (1/cos 2x)(-sin 2x) (2)
y' = (1/cos 2x)(-sin 2x) (2) (y)
=(1/cos 2x)(-sin 2x) (2) (cos 2x)^x
= (-2tan2x)(cos 2x)^x

Answer 3

and that would be (log cos 2x -2 x tan2x )

Answer 4

Sorry, I have some amendment

Answer 5

ln y = x ln cos 2x
diff. both side w.r.t. x
(1/y)(y') = x(1/cos 2x)(-sin 2x) (2) + ln cos 2x
y' =x (1/cos 2x)(-sin 2x) (2) (y) + ln cos 2x (y)
=(x(1/cos 2x)(-sin 2x) (2) +ln cos 2x )(cos 2x)^x
=( x(-2tan2x)(cos 2x)^x +ln cos 2x) (cos 2x)^x

Answer 6

thanks :)

Answer 7

answer was (cos 2x)^x (ln cos 2x-2x tan 2x)