Find y' for y = ln (x+ sqrt(x²+a²)), where a ≠ 0

Find $y'$ for $y = \ln\!\left(x+\sqrt{x^{2}+a^{2}}\right)$, where $a \neq 0$.

Step 1: Differentiate the inner function

With $u = x+\sqrt{x^{2}+a^{2}}$, the chain rule on the radical gives

\[ u' = 1 + \frac{2x}{2\sqrt{x^{2}+a^{2}}} = 1 + \frac{x}{\sqrt{x^{2}+a^{2}}} = \frac{\sqrt{x^{2}+a^{2}}+x}{\sqrt{x^{2}+a^{2}}}. \]

Notice that the numerator of $u'$ is exactly $u$ — this is the coincidence that makes the problem work.

Step 2: Form $y' = \dfrac{u'}{u}$

\[ y' = \frac{1}{x+\sqrt{x^{2}+a^{2}}}\cdot\frac{x+\sqrt{x^{2}+a^{2}}}{\sqrt{x^{2}+a^{2}}} = \frac{1}{\sqrt{x^{2}+a^{2}}}. \]

Step 3: Verification

\[a = 1,\ x = 0:\quad y' = 1; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]

\[a = 1,\ x = 1:\quad y' = \frac{1}{\sqrt2} \approx 0.70711 \;\;\checkmark\]

Step 4: Relation to the inverse hyperbolic sine

Since $\operatorname{arsinh} t = \ln\bigl(t+\sqrt{t^{2}+1}\bigr)$, putting $t = \dfrac{x}{|a|}$ gives

\[ \operatorname{arsinh}\frac{x}{|a|} = \ln\frac{x+\sqrt{x^{2}+a^{2}}}{|a|} = y - \ln|a|. \]

So $y$ and $\operatorname{arsinh}\dfrac{x}{|a|}$ differ by the constant $\ln|a|$ — which is exactly why they have the same derivative.