find the dy/dx of y=log tanx
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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
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
pushpendra926698:
my friend you are to close but your answer is not right
answer will be
2cosec2x
answer is right but anither format you can convert in this format
now you can see that answrr is 100% corrsct
ya i know
but i can sad that i want a proper answer
proper possible only when , if you give which format answrr is given , otherwise not possible ,
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