Question

Does this look right to you guys
y=sinx+sin^2 1
y'=(sinx)(d/dx)+(2sin1)(d/dx)
y'=(sinx)(cos1)+(2sin1)(2cos1)

Answer 1

are you just deriving y as a function? or is this diff eq?

Answer 2

as a function so that i can get a tangent line from it, i think

Answer 3

then no
it should be
y' = cos(x) = 2sin(1)(cos(1))

Answer 4

y' = cos(x) + 2sin(1)(cos(1))
sorry -----^

Answer 5

ok so then i would just plug in my (x,y) which is 0,0 into that formula correct?

Answer 6

if you are trying to find the slope at (0, 0), then yes

Answer 7

alright thanks

Answer 8

no problem