How would I find the least exponent of this problem?

Question

Least exponent, meaning to factor by the lowest exponent. With that said the problem I am having trouble with is...

$$ \mathrm{y}\left(x\right) = \frac{\sqrt{1 - \ln^{2}\left(x\right)} - x\left(-\,{1 \over 2}\right) \left[1 - \ln^{2}\left(x\right)\right]^{-1/2}(-2\ln\left(x\right)/x)} {1 - \ln^{2}\left(x\right)} $$

The only thing that I see that are common in the numerator are the $\sqrt{1 - \ln^{2}\left(x\right)}$ and $\left[1 - \ln^{2}\left(x\right)\right]^{-1/2}$. So then I would factor out the one with the lower exponent which would be the $\left[1 - \ln^{2}\left(x\right)\right]^{-1/2}$ and the reason for that is because $\sqrt{1 - \ln^{2}\left(x\right)}$ becomes $\left[1 - \ln^{2}\left(x\right)\right]^{1/2}$ which has an exponent of $1/2$. The factored out $\left[1 - \ln^{2}\left(x\right)\right]^{-1/2}$ would then go in the denominator because it is negative and that is why the denominator in the answer is $\sqrt{\left[1 - \ln^{2}\left(x\right)\right]^{3}}$. After that, I'm not sure what to do next.

And the answer is supposed to be... $$ \mathrm{y}\left(x\right) = \frac{\left(-1\right)\left[lnx^2x - \ln\left(x\right) - 1\right]} {\sqrt{\left[1 - \ln^{2}\left(x\right)\right]^{3}}} $$

Answer 1

Here is a slightly different approach to simplify the expression.

We obtain \begin{align*} y(x)&=\frac{\sqrt{1-\ln^2x}-x({-\frac12})(1-\ln^2x)^{-\frac12}(-2(\ln x)*\frac1x)}{1-\ln^2x}\\ &=\frac{\sqrt{1-\ln^2x}-\ln x(1-\ln^2x)^{-\frac 12}}{1-\ln^2x}\tag{1}\\ &=\frac{\sqrt{1-\ln^2x}-\ln x(1-\ln^2x)^{-\frac 12}}{1-\ln^2x}\cdot\frac{(1-\ln^2x)^{\frac 12}}{(1-\ln^2x)^{\frac 12}}\tag{2}\\ &=\frac{1-\ln^2 x-\ln x}{\sqrt{(1-\ln^2x)^{3}}}\\ \end{align*}

Comment:

• In (1) we simplify the numerator by doing some cancellation.

• In (2) it is convenient to expand numerator and denominator with $(1-\ln^2x)^{\frac 12}$. Note that \begin{align*} \sqrt{1-\ln^2x}\cdot(1-\ln^2x)^{\frac 12}&=1-\ln^2 x\\ (1-\ln^2x)^{-\frac 12}\cdot(1-\ln^2x)^{\frac 12}&=1 \end{align*}