Find the derivative of ln(x + sqrt(x^2 - a^2)) with respect to x.

Step-by-step solution

1. 1

   Set up the logarithmic chain rule. With $u = x + \sqrt{x^2 - a^2}$ we have $y = \ln u$, so $\dfrac{dy}{dx} = \dfrac{u'}{u}$. The whole problem is now computing $u'$ and hoping it shares a factor with $u$ — which, for this particular combination, it does.

2. 2

   Differentiate the square root. By the chain rule,

   $\frac{d}{dx}\sqrt{x^2 - a^2} = \frac{2x}{2\sqrt{x^2 - a^2}} = \frac{x}{\sqrt{x^2 - a^2}}$

   so $u' = 1 + \dfrac{x}{\sqrt{x^2 - a^2}}$.

3. 3

   Combine the numerator over one denominator.

   $u' = \frac{\sqrt{x^2 - a^2} + x}{\sqrt{x^2 - a^2}}$

   The numerator is now literally $u$ — this is the structural coincidence that makes the answer so clean.

4. 4

   Divide and cancel.

   $\frac{dy}{dx} = \frac{u'}{u} = \frac{x + \sqrt{x^2 - a^2}}{\sqrt{x^2 - a^2}} \cdot \frac{1}{x + \sqrt{x^2 - a^2}} = \frac{1}{\sqrt{x^2 - a^2}}$

5. 5

   Identify the function and its domain. Since $\operatorname{arcosh}(x/a) = \ln\!\left(x + \sqrt{x^2 - a^2}\right) - \ln a$, the expression is the inverse hyperbolic cosine up to an additive constant, which is why its derivative matches $\frac{d}{dx}\operatorname{arcosh}(x/a)$. It is valid for $x > |a|$, where the radicand is positive. Note the contrast with $\ln\!\left(x + \sqrt{x^2 + a^2}\right)$, whose derivative is $1/\sqrt{x^2 + a^2}$ (the arsinh case) and which is defined for all real $x$.

6. 6

   Numerical check. Take $a = 2$, $x = 3$. A central difference with step $10^{-6}$ gives $0.4472136$, and $1/\sqrt{9-4} = 1/\sqrt5 = 0.4472136$. They agree to seven decimals.