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

1 Answer
Jun 26, 2016

Answer:

#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]}