calculus help please I don't understand it. thank you?

Question

find y' (prime) if x=tan(x+y)

Answer 1

You're looking for the first derivative I assume?

x=tan(x+y)
Derive with respect to y, so we get
1=sec^2(x+y) * dy/dx
1/sec^2(x+y) =dy/dx=y'

Note: If f[x]=tanx, then f'[x]=sec^2x * chain

Answer 2

we have ,( tanu) ' = [(sec)^2u]( u ' )
u = x+ y , u ' = 1 + y '
so
tan(x + y) - x = 0
take derivative
[ tan(x + y)] ' - x ' = 0
sec ^2( x + y)( 1 + y ' ) - 1 = 0
sec ^2(x + y) + [sec ^2(x + y)](y ' )] - 1 = 0
sec ^2(x + y)( y ' ) = 1 - sec ^2( x + y)
y ' = [1 - sec ^2(x + y) ] / sec ^2( x + y)

Answer 3

it called implicit differentiate

take dy/dx both sides

1 = sec^2( x+y) *( x+y)'

1= sec^2(x+y)(1 +y')
1 = sec^2(x+y) + y'sec^2(x+y)

1-sec^2(x+y) = y'sec^2(x+y)
[1-sec^2(x+y)]/sec^2(x+y) = y'

answer is y' = [1-sec^2(x+y)]/sec^2(x+y)

or y' = {1/sec^2(x+y)} - (sec^2(x+y)/sec^2(x+y))
by separating factions

y' = cos^2(x+y) -1 if they want answer to be reducing possible.

Answer 4

looks like some sort of curve, maybe a steep learning curve? or an identity crisis.

you have to solve it. for y, in fact.

take it a little at a time. in fact, if you take it in smaller and smaller pieces, you'll have problem with it at all.

p.s. not just chaining you along; this answer is full of hints.