Question

find the derivative of e^(1/x) + 1/(ex^2)
don't quite know how to solve this :/

Answer 1

Hmm this one isn't too bad :)
Ooo a dolphin!

Answer 2

Is this what the problem is supposed to look like?\[\Large\rm e^{1/x}+\frac{1}{e^{x^2}}\]Or this?\[\Large\rm e^{1/x}+\frac{1}{ex^2}\]

Answer 3

Or neither? :U

Answer 4

uh, its the first one @zepdrix

Answer 5

Let's apply our exponent rule to write the expression like this:\[\Large\rm e^{1/x}+e^{-x^2}\]

Answer 6

Do you remember your derivative of e^x?

Answer 7

yes its e^x

Answer 8

mmm ok good.
So e^x gives us e^x.
Gives us the same thing back.

Same idea for more complex exponentials like this one, we simply have to apply the chain rule on top of that.

Answer 9

\[\Large\rm \frac{d}{dx}e^{1/x}=e^{1/x}\left(\frac{d}{dx}\frac{1}{x}\right)\]The derivative of the exponential gave us the same thing back.
Chain rule tells us to multiply by the derivative of the inner function.

Answer 10

If you forgot how to differentiate \(\Large\rm \frac{1}{x}\), you can rewrite it as \(\Large\rm x^{-1}\) and apply your power rule.

Answer 11

okay wait so can you write it out so i can see exactly what you mean by that

Answer 12

I guess you need to remember this with your exponent rules:\[\Large\rm x^{-a}=\frac{1}{x^a}\]The negative tells us to flip it into the denominator.
In our problem, we're doing that in reverse.\[\Large\rm \frac{1}{x^1}=x^{-1}\]

Answer 13

yes i understand that part but are u saying that the e^(1/x) stays the same?

Answer 14

yes, that portion stays the same.
Only need to apply chain rule.

Answer 15

okay so then the answer would be e^(1/x)-2ex^(-3) ?