Y=e^x differentiate using the first principle
4 answers
do you mean the limit of the difference quotient? There are several good discussion of the topic online. Google is your friend.
I mean find dy/dx of dat question using d first priniciple
y (x+h) = e^(x+h) = e^x e^h
y(x+h) - y(x) = e^x e^h - e^x
= e^x (e^h - 1)
so
[y(x+h) - yx) ] / h = e^x [e^h-1]/h
as h ----> 0
e^h ----> 1 + h + h^2/2! + h^3/3! ....
so
as h-->0 e^h -1 -----> h + h^2/2 + h^3/3!...
which is h in the limit
so we have
e^x [ h/h }
which is
e^x
will wonders never cease
y(x+h) - y(x) = e^x e^h - e^x
= e^x (e^h - 1)
so
[y(x+h) - yx) ] / h = e^x [e^h-1]/h
as h ----> 0
e^h ----> 1 + h + h^2/2! + h^3/3! ....
so
as h-->0 e^h -1 -----> h + h^2/2 + h^3/3!...
which is h in the limit
so we have
e^x [ h/h }
which is
e^x
will wonders never cease
It always seems like cheating to use a Taylor Series to prove a derivative...
But nice of you to step in and help. It's clear that google is of no use to someone who cannot spell...
But nice of you to step in and help. It's clear that google is of no use to someone who cannot spell...
0 answers