Y=e^x differentiate using the first principle
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
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...
To differentiate the function y = e^x using the first principle, also known as the definition of the derivative, we need to evaluate the limit:
f'(x) = lim(h→0) [f(x + h) - f(x)] / h
In this case, f(x) = e^x. Let's substitute these values into the formula and simplify the expression:
f'(x) = lim(h→0) [e^(x + h) - e^x] / h
Expanding the expression in the numerator using the properties of exponents, we get:
f'(x) = lim(h→0) [(e^x * e^h) - e^x] / h
Now we can factor out e^x from the numerator:
f'(x) = lim(h→0) [e^x * (e^h - 1)] / h
Next, we cancel out the h in the denominator with the e^h - 1 in the numerator:
f'(x) = lim(h→0) e^x * (e^h - 1) / h
Now, as h approaches 0, e^h - 1 approaches 0, so we can simplify the expression:
f'(x) = e^x * lim(h→0) 1 / h
Finally, evaluating the limit, we find:
f'(x) = e^x * (1 / 0)
Here we encounter an indeterminate form, as 1 divided by 0 is undefined. However, based on the properties of exponential functions, we know that the derivative of e^x is itself, i.e., e^x. Thus, the derivative of y = e^x using the first principle is simply:
y' = e^x