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

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

Mathematics
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are you just deriving y as a function? or is this diff eq?
as a function so that i can get a tangent line from it, i think
then no it should be y' = cos(x) = 2sin(1)(cos(1))

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y' = cos(x) + 2sin(1)(cos(1)) sorry -----^
ok so then i would just plug in my (x,y) which is 0,0 into that formula correct?
if you are trying to find the slope at (0, 0), then yes
alright thanks
no problem

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