Derivative of e^1/x?

Question

Can someone confirm what the derivative of

e^1/x

is, and how it is derived?

Yes, sorry I should have said

e^(1/x)

Answer 1

You have to use proper notation.
Is it (e^1)/x or e^(1/x)?

I'll assume it's the latter.

Using the chain rule, we find the derivative

e^(1/x) multiplied by the derivative
of 1/x

=(-1/x^2)(e^(1/x))

Answer 2

Here, we use the chain rule. We know the derivative of e^u with respect to u is e^u. Since u = 1/x and du/dx =-1/(x^2), it follows that (e^(1/x))' = e^u du/dx = e^(1/x) (-1/x^2) = -e^(1/x)/(x^2), for x <>0.

Answer 3

Hi,

a) d/dx(e^1/x) = e^1/x * d/dx(1/x)
= - (e^1/x)/x^2

b) The derivative of e^x is e^x itself, which means
the slope of the graph e^x i at any point x is the value of e^x at x

The proof follows Picard–Lindelöf theorem which can be seen at the link
http://en.wikipedia.org/wiki/Picard-Lindel%C3%B6f_theorem

Good luck

Answer 4

I am guessing you mean e^(1/x)

Some rules: e^(x) = e^(x)
f(g(x)) = f'(g(x))*g'(x) Chain Rule

f(g(x))=e^(g(x))
g(x)=(1/x) = x^(-1)
g'(x) = -x^(-2)
so
e^(x^(-1))*(-x^(-2)) = -e^(1/x)/(x^2)