Find y' for y = ln (x+ sqrt(1-x²))/x

Find $y'$ for $y = \ln\dfrac{x+\sqrt{1-x^{2}}}{x}$.

Step 1: Split the logarithm

\[ y = \ln\!\left(x+\sqrt{1-x^{2}}\right) - \ln x. \]

Step 2: Differentiate the first logarithm

With $u = x+\sqrt{1-x^{2}}$: \[u' = 1 + \frac{-2x}{2\sqrt{1-x^{2}}} = 1 - \frac{x}{\sqrt{1-x^{2}}} = \frac{\sqrt{1-x^{2}}-x}{\sqrt{1-x^{2}}}.\]

Here the numerator is $\sqrt{1-x^{2}}-x$, which is not the same as $u = x+\sqrt{1-x^{2}}$ — the signs differ. Unlike Problem 122, nothing cancels: $\frac{u'}{u} = \frac{\sqrt{1-x^{2}}-x}{\left(x+\sqrt{1-x^{2}}\right)\sqrt{1-x^{2}}}$.

Step 3: Differentiate the second term and combine

\[ y' = \frac{\sqrt{1-x^{2}}-x}{\left(x+\sqrt{1-x^{2}}\right)\sqrt{1-x^{2}}} - \frac{1}{x}. \]

Step 4: Simplify

Putting the two terms over the common denominator $x\left(x+\sqrt{1-x^{2}}\right)\sqrt{1-x^{2}}$ and expanding the numerator: \[x\left(\sqrt{1-x^{2}}-x\right) - \left(x+\sqrt{1-x^{2}}\right)\sqrt{1-x^{2}} = x\sqrt{1-x^{2}} - x^{2} - x\sqrt{1-x^{2}} - \bigl(1-x^{2}\bigr) = -1.\]

The whole numerator collapses to $-1$, giving

\[ y' = -\frac{1}{x\sqrt{1-x^{2}}\left(x+\sqrt{1-x^{2}}\right)}. \]

Step 5: Verification

\[x = \tfrac12:\quad \sqrt{1-x^{2}} = \frac{\sqrt3}{2} \approx 0.86603,\quad x+\sqrt{1-x^{2}} \approx 1.36603\]

\[y' \approx -\frac{1}{0.5\times 0.86603\times 1.36603} \approx -1.69060\]

\[\frac{y(0.5+10^{-6})-y(0.5-10^{-6})}{2\times 10^{-6}} \approx -1.69060 \;\;\checkmark\]

Step 6: A warning about the false shortcut

It is tempting to imitate Problem 122 and claim that $\dfrac{u'}{u}$ reduces to $\dfrac{1}{\sqrt{1-x^{2}}}$, which would give the plausible-looking answer $\dfrac{1}{\sqrt{1-x^{2}}}-\dfrac{1}{x}$. That is wrong. The step in Problem 122 works because $u'$ has numerator $x+\sqrt{x^{2}+a^{2}} = u$; here the sign flips and the numerator is $\sqrt{1-x^{2}}-x \neq u$. Numerically, at $x = \tfrac12$ the false answer gives

\[ \frac{1}{\sqrt{0.75}} - \frac{1}{0.5} \approx 1.15470 - 2 = -0.84530, \]

which is exactly half the true value $-1.69060$ — close enough to look convincing, and still wrong.