Differentiate to the $1/2$ power Differentiate $ax^2 + b$ to the $1/2$ power. (Exact wording of question)

The answer is $2ax$.

I don't get the question. Is it ${d\over dx} \left(ax^2+b\right)^{1/2}$? If so I am getting 
$$ax\over \sqrt{a x^2+b}$$
It's only when I don't do the half power thing that I get $2ax$.
If we really talk about fractional derivatives, then it is defined as following:$$D^ax^k = \frac{d^a}{dx^a} x^k = \frac{k!}{(k-a)!}x^{k-a} = \frac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a}$$So using $\Gamma$ function we can define fractional derivative. It has all the properties of linear operator and is an extension of regular derivative. For example for $n = a+b$: $$D^aD^bx^k\frac{d^a}{dx^a} \frac{d^b}{dx^b}x^k = \frac{d^n}{dx^n}x^k = D^nx^k$$. To make sure it is consistent with regular derivative we set that fractional derivative of constant is zero, i.e. $D^a(const) = 0$. Then $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x^k = \frac{\Gamma(k+1)}{\Gamma(k+\frac{1}{2})}x^{k-\frac{1}{2}}$$For $k=2$$$D^{\frac{1}{2}}x^2 = \frac{2}{\frac{3}{4}\sqrt{\pi}}x\sqrt{x}$$ So $$D^{\frac{1}{2}}(ax^2+b) = \frac{8}{3\sqrt{\pi}}ax\sqrt{x}$$
If we really talk about fractional derivatives, then it is defined as following:$$D^ax^k = \frac{d^a}{dx^a} x^k = \frac{k!}{(k-a)!}x^{k-a} = \frac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a}$$So using $\Gamma$ function we can define fractional derivative. It has all the properties of linear operator and is an extension of regular derivative. For example for $n = a+b$: $$D^aD^bx^k\frac{d^a}{dx^a} \frac{d^b}{dx^b}x^k = \frac{d^n}{dx^n}x^k = D^nx^k$$. To make sure it is consistent with regular derivative we set that fractional derivative of constant is zero, i.e. $D^a(const) = 0$. Then $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x^k = \frac{\Gamma(k+1)}{\Gamma(k+\frac{1}{2})}x^{k-\frac{1}{2}}$$For $k=2$$$D^{\frac{1}{2}}x^2 = \frac{2}{\frac{3}{4}\sqrt{\pi}}x\sqrt{x}$$ So $$D^{\frac{1}{2}}(ax^2+b) = \frac{8}{3\sqrt{\pi}}ax\sqrt{x}$$