1. ## typo checking ....

I am asked to differentiate $y = 2x^x$

taking logs to base e

$\ln y = x \ln 2x$

implicit differentiation

$\dfrac{1}{y} \dfrac{dy}{dx} = x \dfrac{2}{2x} + 1. \ln 2x$

$\dfrac{1}{y} \dfrac{dy}{dx} = 1 + \ln 2x$

$\dfrac{dy}{dx} = 2x^x (1 + \ln 2x)$

I think this is the right answer. The answer given is $2x^x ( 1 + \ln x)$ am I doing something wrong is it a mistake in the answer?

2. ## Re: typo checking ....

Your first step is incorrect. If $\displaystyle y= 2x^x$ then $\displaystyle log(y)= log 2+ x log(x)$. The "2" is NOT raised to the x power.