Exercise 5.3 in Calculus Made Easy, by Silvanus Thompson, is to find $\mathrm d\over \mathrm dx$ when the following relationship holds ($a$ and $b$ are both constants):

$$ay + bx = by - ax + (x + y)\sqrt{a^2 - b^2}$$

I tried differentiating both sides of this equation with respect to $x$ as follows:

$$\begin{align} {\mathrm d\over \mathrm dx}(ay + bx) &= {\mathrm d\over \mathrm dx}\left(by - ax + (x + y)\sqrt{a^2 - b^2}\right)\\ a{\mathrm dy\over \mathrm dx} + b &= b{\mathrm dy\over \mathrm dx} - a + \left(1 + {dy\over dx}\right)\sqrt{a^2 - b^2}\\ \left(a - b - \sqrt{a^2 - b^2}\right){\mathrm dy\over \mathrm dx} &= \sqrt{a^2 - b^2} - a - b\\ {\mathrm dy\over \mathrm dx} &= {\sqrt{a^2 - b^2} - a - b \over a - b - \sqrt{a^2 - b^2}} \end{align}$$

The author has considerably more finess and first does some algebraic manipulation to get the equation into better shape. First he squares both sides, then pushes some things around, then takes the square root of both sides:

$$\begin{align} (a - b)^2y^2 + (a + b)^2x^2 + 2(a + b)(a - b)xy &= (x^2 + y^2 + 2xy)(a^2 - b^2)\\ \left[(a - b)^2 - \left(a^2 - b^2\right)\right]y^2 &= \left[\left(a^2 - b^2\right) - (a + b)^2\right]x^2\\ 2b(b - a)y^2 &= -2b(b + a)x^2\\ y &= \sqrt{a + b \over a - b}x\\ {\mathrm dy\over \mathrm dx} &= \sqrt{a + b \over a - b} \end{align}$$

I believe that both answers are correct, but if that is true, it should be possible to convert one to the other algebraically. For the life of me, I can't see how to do this. So my question is: are these two answers really equivalent?

  • 2
    $\begingroup$ Note that if you had done the same work before differentiating to isolate $y$ that you did afterwards to isolate $\frac{dy}{dx}$, you would have found $$y = {\sqrt{a^2 - b^2} - a - b \over a - b - \sqrt{a^2 - b^2}}x$$which at least tells you the issue was strictly algebraic. Note also that Dr. Thompson can only get away with his manipulations because he assumes that $x$ and $y$ are positive. Otherwise his method would require additional care that he ignores. $\endgroup$ May 9, 2017 at 17:10

1 Answer 1


$${\sqrt{a^2 - b^2} - a - b \over a - b - \sqrt{a^2 - b^2}}=\frac{\sqrt{a+b}\left(\sqrt{a-b}-\sqrt{a+b}\right)}{\sqrt{a-b}\left(\sqrt{a-b}-\sqrt{a+b}\right)}=\frac{\sqrt{a+b}}{\sqrt{a-b}}$$

  • $\begingroup$ Wow! How did you "see" that? $\endgroup$
    – Alex D
    May 9, 2017 at 13:46
  • 2
    $\begingroup$ Given the desired endpoint, it was probably just a matter of seeing whether you can factor the desired numerator out of the top, and the desired denominator out of the bottom. Turns out, that worked in this case. $\endgroup$ May 9, 2017 at 14:01
  • 2
    $\begingroup$ @AlexD Also, you have a Difference of Two Squares ($a^2-b^2$) which is one of the "I need to factor this" situations. $\endgroup$ May 9, 2017 at 15:50

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