Differentiate $f(x) = x \tan^{-1}\sqrt{x}$
Solution: $$f'(x) = x \frac{1}{{1+(\sqrt{x}})^2} \frac{1}{2}x^{-1/2} + \tan^{-1}{\sqrt{x}} = \frac{\sqrt{x}}{2(1+x)} + \tan^{-1}\sqrt{x}$$
I see the chain rule is being applied to $\sqrt{x}$ in the denominator, resulting in $\frac{1}{2}x^{-1/2}$ however, the solution has $\sqrt{x}$ in the numerator.
My question: Why does the $\sqrt{x}$ move to the numerator? The differentiation takes place on a function in the denominator which results in a negative exponent. I understand negative exponents in the denominator move to the numerator.
My (edited) question: Why does $\frac{1}{2}x^{-1/2}$ not equal $\frac{1}{2\sqrt{x}}$?
Update: answered.
You forgot the $x$ term multiplying the troublesome expression (from the product rule). So you have $$ x\frac{1}{1+\sqrt{x}^2}\frac{1}{2}x^{-1/2}=\frac{\sqrt{x}}{2}\frac{1}{1+\sqrt{x}^2} $$