y= (1-tan x)/sec x
= (1-sinx/cosx)*cosx=cosx-sinx
y'=-sinx-cosx
y=(1+tan3x)^3/2
y'=3/2(1+tan3x)^1/2 (3sec^2 3x)
y= (1-tan x)/sec x
find dy/dx
y= (1 + tan3x) ^(3/2)
find dy/dx
thanks a lot
3 answers
for the question
y= (1 + tan3x) ^(3/2)
find dy/dx
the ans is
y'=9/2(1+tan3x)^1/2 (3sec^2 3x)
y= (1 + tan3x) ^(3/2)
find dy/dx
the ans is
y'=9/2(1+tan3x)^1/2 (3sec^2 3x)
bob is right,
mrfrank has an extra 3 hanging around
should be
(9/2)(1+tan (3x))^(1/2) (sec^2 (3x) )
mrfrank has an extra 3 hanging around
should be
(9/2)(1+tan (3x))^(1/2) (sec^2 (3x) )