Differentiate the Function: $y=x^x$ $y=x^x$
Use $\frac{d}{dx}(a^x)=a^x \ln a$
My answer is: $x^x \ln x$
The book has the answer as $x^x\ (1+ \ln\ x)$
Am I missing a step? 
 A: If $f(x) = x^x$ then taking the natural logarithm of both sides and making use of the power rule for logarithms yields $$\ln f(x) = x \ln x$$ Implicitly differentiating with respect to $x$ gives us $$\frac{f'(x)}{f(x)} = \ln x + 1$$
So, by multiplying through by $f(x)$, we have $$\bbox[10px, border: blue 1px solid]{f'(x) = x^x(\ln x + 1).}$$
The issue with your method is that $\frac{\mathrm{d}}{\mathrm{d}x}a^x = a^x \ln a$ holds only when $a$ is constant, in this case we have $a$ as some variable $x$.
A: Take logs first so that you have $$\ln y =x\ln x$$
Then $$\frac 1y \frac{dy}{dx}=1+\ln x$$
And the result follows
A: $(x^x)'=(e^{x\ln x})'=x^x(1\ln x+x\frac1x)$.
Incidentally, if we add (wrong) solutions $(x^x)'=(a^x)'=a^x\ln a$ and $(x^x)'=(x^a)'=a \cdot x^{a-1}$ for $a=x$, we obtain the correct result. :-)
A: taking the logarithm of both sides we get
$$\ln(y)=x\ln(x)$$ differentiating with respect to $x$ we obtain
$$\frac{1}{y}y'=\ln(x)+1$$ multiplying by $y$ you will get the result.
A: $\bf{My\; Solution::}$ Let $y=x^x\;,$ Now Taking $\log_{e}$ on both side, we get
$$\displaystyle \log_{e}(y) = \log_{e}(x)^x = x\cdot \log_{e}(x)$$
Now Differentiate both side $\bf{w.r.t}$ $x\;,$ We get
$$\displaystyle \frac{1}{y}\cdot \frac{dy}{dx} = x\cdot \frac{1}{x}+\log_{e}(x)\cdot 1\Rightarrow \frac{dy}{dx} = y(1+\log_{e}(x)) = x^x\cdot (1+\log_{e}(x)).$$ 
A: 
Am I missing a step ?

Of course ! You are missing $(x^n)'=n~x^{n-1}$. Add this to your initial incomplete result, then replace n by x — just like you also replaced a by x — and see what you get ! ;-$)$
