If (y=x^{e^x}+x^{e} for x>0),the ( \dfrac{dy}{dx}) is equal... | Mathematics | SolveeIt

If $y=x^{e^x}+x^{e} for x>0$ ,the $\dfrac{dy}{dx}$ is equal to ?

$x^{e^x}[1/x+lnx]+e^x$

$x^{e^x} e^x[1/x+lnx]+ex^e-1$

$e^x.x^{e^x-1}+ex^e$

$x^{e^x }e^-x[1/x-lnx]+ex^e-1$

$x^{e^x} e^x[1/x-lnx]+ex^e-1$

$=e^{x}.x^{e^x-1}+e.x^{e-1}$

Answer

$=e^{x}.x^{e^x-1}+e.x^{e-1}$

Explanation

$y=x^{e^x}+x^{e} \text{ for } x>0$

Now $\dfrac{dy}{dx}$ for $x>0$ can be represented as

$\dfrac{d}{dx}(x^{e^x}+x^{e})$

$=e^{x}.x^{e^x-1}+e.x^{e-1}$ (_Ans)

Mathematics Question on Differential Calculus