f(x)=2 ln [cosx]?

0 ⤊

f(x)=2 ln [cosx]

what is f''(x)=?

2007-11-01 14:15:34 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

f(x)=2ln[cos(x)]

By chain rule:
f'(x) = 2 * (1 / cos(x)) * -sin(x)
= -2sin(x)/cos(x)
= -2tan(x)

2007-11-01 14:20:09 · answer #1 · answered by whitesox09 7 · 0⤊ 0⤋

Use the chain rule.
d/dx(ln(x))=1/x
d/dx(cos(x))=-sin(x)

so
f'(x) = 2*(1/cos(x))*(-sin(x))
f'(x)=-2*tan(x)

2007-11-01 14:37:41 · answer #2 · answered by Not Eddie Money 3 · 0⤊ 0⤋

f´(x) = 2/cos(x) * (-sin x)

2007-11-01 14:20:35 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋