Find dy/dx through implicit differentiation: sin(x)=x(1+tan(y))

iroroPagbublh

iroroPagbublh

Answered question

2022-08-11

Find d y / d x through implicit differentiation:
sin ( x ) = x ( 1 + tan ( y ) )
My solution
(1) d d x [ sin ( x ) ] = d d x [ x ( 1 + tan ( y ) ) ] (2) cos ( x ) = ( 1 ) ( 1 + tan ( y ) ) + x ( 1 + tan ( y ) ) 1 ( sec 2 ( y ) ) d y d x (3) cos ( x ) = ( 1 + tan ( y ) ) + x ( sec 2 ( y ) ) 1 + tan ( y ) d y d x (4) cos ( x ) ( 1 + tan ( y ) ) = x ( sec 2 ( y ) ) 1 + tan ( y ) d y d x (5) d y d x = ( cos ( x ) 1 tan ( y ) ) ( 1 + tan ( y ) ) x ( sec 2 ( y ) )
Solution from manual I'm using
(6) sin ( x ) = x ( 1 + tan ( y ) ) (7) cos ( x ) = x ( s e c 2 ( y ) ) y + ( 1 + tan ( y ) ) ( 1 ) (8) y = cos ( x ) tan ( y ) 1 x sec 2 ( y )

Answer & Explanation

Pasrbekwp

Pasrbekwp

Beginner2022-08-12Added 10 answers

Your step (2) is incorrect; the product rule states that
d d x x ( 1 + tan y ) = d ( x ) d x ( 1 + tan y ) + x d ( 1 + tan y ) d x = ( 1 + tan y ) + x ( 0 + sec 2 y d y d x )
cortejosni

cortejosni

Beginner2022-08-13Added 5 answers

d d x [ x ( 1 + tan ( y ) ) ] = ( x ) ( d d x [ 1 + tan ( y ) ] ) + ( 1 ) ( 1 + tan ( y ) ) = ( x ) ( 0 + sec 2 ( y ) d y d x ) + ( 1 + tan ( y ) ) = x sec 2 ( y ) d y d x + 1 + tan ( y )
Essentially, you differentiated incorrectly by introducing the ( 1 + tan ( y ) ) 1 term. I'm not entirely sure where that came from.

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