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Givenƒ(x) = e^x, verify that lim (e^x+h-e^h) / h = e^x h->O = and explain how this illustrates that ƒ′(x) = ln e • ƒ(x) = ƒ(x)Asked by Anonymous
Given ƒ(x) = e^x, verify that lim h→0 (e^(x+h)−e^x)/h = e^x and explain how this illustrates that ƒ′(x) = ln e • e^x = e^x.
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Answered by
bobpursley
first, remember that e^(a+b)=e^a * e^b
Lim (e^(x+dx) - e^x)/dx)
lim (e^x*e^dx-e^x)/dx=
lim (e^x(e^dx-1)/dx)
e^x lim(e^dx-1/dx)=e^x see https://www.youtube.com/watch?v=qWw8VnzTddg for why.
Lim (e^(x+dx) - e^x)/dx)
lim (e^x*e^dx-e^x)/dx=
lim (e^x(e^dx-1)/dx)
e^x lim(e^dx-1/dx)=e^x see https://www.youtube.com/watch?v=qWw8VnzTddg for why.
Answered by
Anonymous
i already got the same distance as you in finding the derivative but how do i continue from this? the video is a bit confusing on how he made y=e^h-1
Answered by
bobpursley
y is defined as e^h -1, then he explores the rest as a funcion of y.
Answered by
Anonymous
okay but for the second part (explain how this illustrates that ƒ′(x) = ln e • e^x = e^x.) how do i explain it??
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