Question

Can anyone please explain why say dy/dx of e^n gives only ne^n not ne^n-1, or other exponential functions. I've seen alot of vids that just state this rule, which is just no help.
thanks if you can help explain :)

Answer 1

The rule you're thinking of is the power rule.

When you have \(x^n\), the exponent is constant. When you have \(e^x\), you have a function of x as exponent. The same rules don't apply.

Answer 2

As a side not, \(\dfrac{d}{dx}e^n=0\), since \(e^n\) is not a function of \(x\), but I know what you meant :)

Answer 3

@sirshelley If you have
\[\frac{d}{dx} (x^n)=n\times x^{n-1}
\]
But for exponential function
\[\frac{d}{dx} e^x =e^x\]
one more case, when the power of e is x^n\[\frac{d}{dx} e^{x^n} =e^{x^{n}}\times n \times x^{n-1}\]

Answer 4

yeah lol, was just an example , many thanks for clearing that up for me though :)

Answer 5

sorry I didn't mean e^n meant n^e^n , brain, not working :c, gj both on guessing what i meant though :}

Answer 6

..i did it again didn't i lols

Answer 7

You have confusion with this
\[\large n^{e^n}\]

Answer 8

yeah I do, 5am so this is mostly likely why, I do follow though, so many thanks both :)

Answer 9

\[x^{e^x}\]
\[\frac{d}{dx} x^{e^x}=e^x \times x^{e^x-1}\frac d {dx}e^x\]
\[=> e^x \times x^{e^x-1}\times e^x\]
Do you get this?

Answer 10

yeah , I follow :)
That all makes sense, the problem I had was applying the diff rules with exponentials. But you've shown me how it it works out so ,all sorted :)

Answer 11

Cool, :D

Answer 12

Here's a proof of why the derivative of the exponential function, e^x, is equal to e^x.





So, taken together, this means that the derivative of y(x) = e^x is e^x * (slope of e^x at x = 0). The slope of e^x at x = 0 is, by definition, equal to 1. Thus, y'(x) = e^x.