Differentiate y = log (tan x)
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Answer:
y = log tanx
we know, if any function y = logf(x) is given then, dy/dx = 1/f(x).df(x)/dx
use this concept here ,
y = log tanx
differentiate wrt x
dy/dx = 1/tanx .d(tanx)/dx
= 1/tanx .sec²x
= sec²x/tanx
[ sec²x = 1 + tan²x use this ]
dy/dx = (1 + tan²x)/tanx
= 1/tanx + tan²x/tanx
= cotx + tanx
= sinx/cosx + cosx/sinx
= (sin²x + cos²x )/sinx.cosx
= 1/sinx.cosx
= 2/2sinx.cosx
[ 2sinx.cosx = sin2x ]
= 2/sin2x
= 2cosecx
hence, dy/dx = 2cosecx
This is your answer.
Hope this helps you!
Step-by-step explanation:
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