Math Help - Derivative of x^ (e^x)

1. 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 )]

2. Re: Derivative of x^ (e^x)

$x^{e^x}=e^{\ln(x)e^x}$

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

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

3. 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?

4. Re: Derivative of x^ (e^x)

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

$\dfrac d {dx} x^{e^x} \neq x^{e^x - 1}$

Thank you!

6. 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.

7. Re: Derivative of x^ (e^x)

Great, thanks for the clarification Sip