Find y' for y = sqrt(x²+1) + ln (1- sqrt(x²+1))/x

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

Step 1: Introduce a shorthand and split the logarithm

Write $s = \sqrt{x^{2}+1}$, so that $s' = \dfrac{x}{s}$. Splitting the logarithm formally, $y = s + \ln(1-s) - \ln x$.

Step 2: Differentiate each term

\[ \frac{d}{dx}(s) = \frac{x}{s}, \qquad \frac{d}{dx}\ln(1-s) = \frac{-s'}{1-s} = \frac{-x}{s(1-s)}, \qquad \frac{d}{dx}(-\ln x) = -\frac{1}{x}. \]

Step 3: Combine the first two terms

Both carry the factor $\dfrac{x}{s}$: \[\frac{x}{s} - \frac{x}{s(1-s)} = \frac{x}{s}\!\left(1-\frac{1}{1-s}\right) = \frac{x}{s}\cdot\frac{-s}{1-s} = \frac{-x}{1-s}.\]

so that

\[ y' = \frac{-x}{1-s} - \frac{1}{x}. \]

Step 4: Rationalise

Multiply the first term by $\dfrac{1+s}{1+s}$ and use $1-s^{2} = 1-\bigl(x^{2}+1\bigr) = -x^{2}$: $\frac{-x(1+s)}{1-s^{2}} = \frac{-x(1+s)}{-x^{2}} = \frac{1+s}{x}$.

Therefore \[y' = \frac{1+s}{x} - \frac{1}{x} = \frac{s}{x} = \frac{\sqrt{x^{2}+1}}{x}.\]

Step 5: Verification

\[x = -1:\quad s = \sqrt2 \approx 1.41421,\ \frac{1-s}{x} \approx 0.41421,\ y \approx 1.41421 - 0.88137 = 0.53284\]

\[y'(-1) = \frac{\sqrt2}{-1} \approx -1.41421; \qquad \frac{y(-1+10^{-5})-y(-1-10^{-5})}{2\times 10^{-5}} \approx -1.41421 \;\;\checkmark\]

Step 6: Domain

Since $\sqrt{x^{2}+1} \geq 1$, the numerator $1-\sqrt{x^{2}+1}$ is negative for every $x \neq 0$. The quotient inside the logarithm is therefore positive only when $x \lt 0$, so the function is defined exactly on $(-\infty,0)$. Note that the split $\ln\frac{A}{B} = \ln A - \ln B$ used in Step 1 is formal here, since $A$ and $B$ are both negative; the derivative it produces is nevertheless correct, as the numerical check confirms.