Solving for the Tangent Line at (1,e^-2)

[hvroegindewey]

Does anyone know how to find the equation of a tangent line to y=e^(-2x) at the point (1,e^-2) I honestly have no idea how to even start this problem when I tried it I came up with y=1(x+.27)+ln2

 

[reef]

You have to differentiate y=e^(-2x) then solve for f'(1) which will give you the slope. You can solve the equation of the tangent line by plugging values into y-y1=m(x-x1).

Hint: d/du(e^u)=e^u*u'