Find both first partial derivatives of z = ln( (x - sqrt(x^2 - y^2)) / (x + sqrt(x^2 - y^2)) ).

Step-by-step solution

1. 1

   Break the logarithm of a quotient into a difference of logarithms first. Simplifying before differentiating saves an entire quotient-rule computation. Writing $u=\sqrt{x^2-y^2}$,

   $z=\ln(x-u)-\ln(x+u)$

   Also record the two partials of the inner radical once, since both derivatives need them:

   $u_x=\frac{x}{\sqrt{x^2-y^2}}=\frac{x}{u},\qquad u_y=\frac{-y}{\sqrt{x^2-y^2}}=-\frac{y}{u}$

   One more relation will do most of the work later: $x^2-u^2=x^2-(x^2-y^2)=y^2$.

2. 2

   Differentiate with respect to $x$, treating $y$ as a constant. The chain rule on each logarithm gives

   $\frac{\partial z}{\partial x}=\frac{1-u_x}{x-u}-\frac{1+u_x}{x+u}=\frac{1-\dfrac{x}{u}}{x-u}-\frac{1+\dfrac{x}{u}}{x+u}$

   Do not stop here and guess that the two symmetric-looking pieces cancel — they do not. Each one must be simplified first.

3. 3

   Simplify each piece separately. Put the numerators over $u$:

   $\frac{1-\frac{x}{u}}{x-u}=\frac{\frac{u-x}{u}}{x-u}=\frac{-(x-u)}{u(x-u)}=-\frac{1}{u}$

   $\frac{1+\frac{x}{u}}{x+u}=\frac{\frac{u+x}{u}}{x+u}=\frac{x+u}{u(x+u)}=\frac{1}{u}$

   So the first term is $-\tfrac1u$ and the second term is $+\tfrac1u$, and the derivative is their difference:

   $\frac{\partial z}{\partial x}=-\frac{1}{u}-\frac{1}{u}=-\frac{2}{u}=-\frac{2}{\sqrt{x^2-y^2}}$

   This is the step where a sign slip is fatal: subtracting $+\tfrac1u$ from $-\tfrac1u$ gives $-\tfrac2u$, not $0$.

4. 4

   Differentiate with respect to $y$, treating $x$ as a constant. Now only $u$ depends on $y$:

   $\frac{\partial z}{\partial y}=\frac{-u_y}{x-u}-\frac{u_y}{x+u}=\frac{y/u}{x-u}+\frac{y/u}{x+u}=\frac{y}{u}\left(\frac{1}{x-u}+\frac{1}{x+u}\right)$

   Both terms end up positive because each of them carries the same factor $-u_y=+\dfrac{y}{u}$: the minus signs coming from the quotient rule and from $u_y=-y/u$ cancel each other.

5. 5

   Combine the two fractions using $x^2-u^2=y^2$.

   $\frac{1}{x-u}+\frac{1}{x+u}=\frac{(x+u)+(x-u)}{x^2-u^2}=\frac{2x}{y^2}$

   Therefore

   $\frac{\partial z}{\partial y}=\frac{y}{u}\cdot\frac{2x}{y^2}=\frac{2x}{y\,u}=\frac{2x}{y\sqrt{x^2-y^2}}$

6. 6

   Check both answers numerically. Take $x=3$, $y=2$, so $u=\sqrt{9-4}=\sqrt5\approx2.2360680$. A central difference with step $10^{-6}$ gives

   $z_x\approx-0.8944272,\qquad z_y\approx1.3416408$

   and the formulas give $-2/\sqrt5=-0.8944272$ and $2(3)/(2\sqrt5)=1.3416408$. Both match to seven decimals. (A frequently seen wrong answer, $z_x=0$, fails this test immediately: $z$ genuinely does change as $x$ moves with $y$ held fixed.)