find the tangent to the curve at the point corresponding to the given value of the parameter

x=e^(√t)

y= t - ln tē

t = 1

I found...

dx/dt = (e√x)/(2x)

dy/dt = 1 - (2t/tē)

then you do

dy/dt / dx/dt

I got -e from that. I plugged in my coordinates and got

y-1 = -e(x-e)

the book says -(2/e)x+3. How am I so off?

Thanks guys. and no offense but it'd be cool if you could do an entire "debug" in one post instead of asking questions that I probly couldn't answer. Not trying to sound like a jerk but i'm just not the sharpest calc student on the planet so theres no point in wasting our time.