# Derivative of x^ (e^x)

• Jun 12th 2014, 03:58 PM
sakonpure6
Derivative of x^ (e^x)
Why is the answer Wolfram|Alpha Widgets: "Derivative Calculator" - Free Mathematics Widget and not [e^x] [ x^(e^x -1 )]
• Jun 12th 2014, 04:39 PM
romsek
Re: Derivative of x^ (e^x)
$x^{e^x}=e^{\ln(x)e^x}$

$\dfrac d {dx} e^{\ln(x)e^x}=e^{\ln(x)e^x}\dfrac d {dx} \left(\ln(x)e^x\right)=$

$e^{\ln(x)e^x}\left(\dfrac {e^x}{x} + \ln(x)e^x\right)=x^{e^x}\left(\dfrac {e^x}{x} + \ln(x)e^x\right)$
• Jun 12th 2014, 05:06 PM
sakonpure6
Re: Derivative of x^ (e^x)
So we can't take the derivative using the power rule and chain rule for x^(e^x) without re writing it?
• Jun 12th 2014, 05:12 PM
romsek
Re: Derivative of x^ (e^x)
Quote:

Originally Posted by sakonpure6
So we can't take the derivative using the power rule and chain rule for x^(e^x) without re writing it?

no sir

$\dfrac d {dx} x^{e^x} \neq x^{e^x - 1}$
• Jun 12th 2014, 05:54 PM
sakonpure6
Re: Derivative of x^ (e^x)
Thank you!
• Jun 12th 2014, 06:27 PM
SlipEternal
Re: Derivative of x^ (e^x)
The power rule can only be used when the exponent of $x$ is a real number. Since you have the exponent of $x$ is a function of $x$, the power rule doesn't work. Romsek's method works well, as does implicit differentiation:

$y = x^{e^x}$

$\ln y = \ln x^{e^x} = e^x\ln x$

$\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{e^x}{x}+e^x\ln x$

$\dfrac{dy}{dx} = y\left(\dfrac{e^x}{x}+e^x\ln x\right)$

Plugging in for $y = x^{e^x}$ gives:

$\dfrac{dy}{dx} = x^{e^x}\left(\dfrac{e^x}{x}+e^x\ln x\right)$, just as romsek computed.
• Jun 12th 2014, 06:31 PM
sakonpure6
Re: Derivative of x^ (e^x)
Great, thanks for the clarification Sip