What are the first and second derivatives of # g(x) = lncot^2x#?

Question

1842 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-31T19:36:20

see below

#g(x)=lncot^2x#

#g(x)=ln((cotx)^2)#

Use chain rule #(f(g(x)))'=f'(g(x)) * g'(x)#

#g'(x)=1/cot^2x * 2cot x * -csc^2x#

#g'(x)=(-2csc^2x)/cotx = -2csc^2 x tan x#

Use product rule#(fg)'=fg'+gf'# to find #g''(x)#

#f = -2csc^2x, g=tanx#

#f'=-4cscx * -cscx cot x, g'(x)=sec^2x#

#g''(x)=-2csc^2x sec^2x+4csc^2x cot x tanx#

#g''(x)=-2csc^2x sec^2x+4csc^2x#