Question

If y=tan(x+y), then find dy/dx

Answer 1

Answer:

sec² ( x + y ) / ( 1 - sec² ( x + y )

Step-by-step explanation:

Given :

y = tan ( x + y )

Diff. w.r.t. x :

d y / d x = sec² ( x + y ) . ( x + y )'

= > y' = sec² ( x + y ) . ( ( x )' + ( y )' )

= > y' = sec² ( x + y ) . ( 1 + y' )

= > y' = sec² ( x + y ) + y' ( sec² ( x + y ) )

= > y' - y' ( sec² ( x + y ) ) = sec² ( x + y )

= > y' ( 1 - sec² ( x + y ) ) = sec² ( x + y )

= > y' = sec² ( x + y ) / ( 1 - sec² ( x + y )

Therefore , d y / d x of given function y = tan ( x + y ) is sec² ( x + y ) / ( 1 - sec² ( x + y )

Answer 2

Answer:

sec² ( x + y ) / ( 1 - sec² ( x + y )

Step-by-step explanation:

Given :

y = tan ( x + y )

Diff. w.r.t. x :

d y / d x = sec² ( x + y ) . ( x + y )'

= > y' = sec² ( x + y ) . ( ( x )' + ( y )' )

= > y' = sec² ( x + y ) . ( 1 + y' )

= > y' = sec² ( x + y ) + y' ( sec² ( x + y ) )

= > y' - y' ( sec² ( x + y ) ) = sec² ( x + y )

= > y' ( 1 - sec² ( x + y ) ) = sec² ( x + y )

= > y' = sec² ( x + y ) / ( 1 - sec² ( x + y )

Therefore , d y / d x of given function y = tan ( x + y ) is sec² ( x + y ) / ( 1 - sec² ( x + y )