Question

How do you differentiate `y= e^(1-x)`?

Answer 1

`y' = - e^(1-x)`

Explanation:

`y=e^(1-x) = e*e^(-x).`
`y' = (e*e^(-x)) '=e (e^(-x)) ' = e(-e^(-x))`
`= -( e*e^(-x))`
`= -( e^(1-x))`

or using logs

`y=e^(1-x)`
`ln y=1-x`
`1/y y' = -1`
`y' = -y = - e^(1-x)`

slopes okay.
45° at y' = 1 looks okay.
`y'-2.718281828*((2.718281828)^((x)^-2))`.
areas don't look okay

graph{(y-2.718281828/(2.718281828^x))=0 [-1, 4, -0.5, 2]}.

graph{(y-2.718281828*((2.718281828)^x^(-2)))=0 [-1, 4, -0.5, 2]}