How do you find #dy/dx# by implicit differentiation of #xcosy=1# and evaluate at point (2, pi/3)?

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Answer 1 · 2017-03-01T10:05:29

#[dy/dx]_(2,pi/3) = 1/(2sqrt3)#

Differentiate the equation with respect to #x#:

#d/dx (xcosy) = 0#

#cosy -xsinydy/dx = 0#

#cosy = xsinydy/dx #

#dy/dx = 1/(xtgy)#

In the point #(2,pi/3)# the value is then:

#[dy/dx]_(2,pi/3) = 1/(2tg(pi/3)) = 1/(2sqrt3)#