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Hi,

We all know the first derivative of e^(x^2) is 2xe^(x^2). But how are we to find it using the first derivative. I hope anyone can help me. Thanks a lot!

2006-08-18 18:10:00 · 7 answers · asked by sky_blue 1 in Science & Mathematics Mathematics

Thanks for all solutions!

I am sorry but I am trying to find the derivative using the following formula:
f'(x) = lim [ f(x +h) - f(x) ] / h
h->0

Can anyone show me how to find using the formula above? Thanks

2006-08-18 18:33:47 · update #1

7 answers

Since we can't do ascii art I will write e^q as exp(q) for clarity.

f'(x) = lim (h->0) [ f(x +h) - f(x) ] / h

f'(x) = lim(h->0) [ exp( (x+h)^2 ) - exp(x^2) ] / h
= lim(h->0) [ exp( x^2 + 2xh + h^2 ) - exp( x^2 )] / h
= lim(h->0) [ exp( x^2) * exp( 2xh ) * exp( h^2 ) - exp( x^2 ) ] / h
= lim(h->0) [ exp(x^2) * [ exp(2xh)*exp(h^2) - 1 ] ] / h
= lim(h->0) [ exp(x^2) * [ exp(2xh + h^2) - 1 ] ] / h
= exp(x^2) * lim(h->0) [ exp(2xh + h^2) - 1 ] / h

Since the top and bottom of this limit now both go to zero we can use L'Hopital's rule.

= exp(x^2) * lim(h->0) [ (2x + 2h) * exp( 2xh + h^2 ) ] / 1

Clearly now the limit goes to (2x + 0) * exp(0) = 2x * 1 = 2x. Therefore:

= exp(x^2) * 2x which is your answer.

2006-08-18 18:40:54 · answer #1 · answered by selket 3 · 1 0

Divide this expression into recognisable functions.
Let the function becomes e^u and u = x^2
We know that d(e^u) / dx = e^u du/dx.
Substitute x^2 on u.
e^(x^2) d(x^2)/dx = e^(x^2) * (2x) = 2xe^(x^2)

2006-08-19 01:26:35 · answer #2 · answered by Joe Mkt 3 · 0 0

f'(x) = lim [ f(x +h) - f(x) ] / h
h->0

e^((x+h)^2) - e^x^2 = e^x^2 + e^2xh + e^h^2 - e^x^2
= e^2xh + e^h^2

USE the following standardlimit :
lim (n -> INF)(1+x/n)^n=e^x.

set n = 1/h and note that lim h->0 e^h^2/h = 1

etc

2006-08-19 01:45:31 · answer #3 · answered by gjmb1960 7 · 0 0

The derivative of exp(x) is exp(x), but it can also be written exp(x)*f'(x) = exp(x)*1

now if you were to take the derivative of exp(x^2) it would be exp(x^2)*f'(x), in this case f'(x) = 2x, therefore, the derivative of exp(x^2) is

exp(x^2)*2x

2006-08-19 01:23:37 · answer #4 · answered by gtn 3 · 0 0

This is the chain rule:

g(x)=x^2
f(x)= e^x

f(g(x)) = e^(x^2)

Using the chain rule for differentiation, you arrive at the answer.

2006-08-19 01:22:00 · answer #5 · answered by a_liberal_economist 3 · 0 0

e^2x

My friend's gave me the answer...I have no idea what this means.

He told me to tell you the jerk is zero.

2006-08-19 01:16:59 · answer #6 · answered by Kimbie 1 · 0 0

eww.. i hate calculus

2006-08-19 01:18:58 · answer #7 · answered by Anonymous · 0 0

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