<s>[INST] sum of legs of inscribed right triangle Consider a right and isoceles triangle ABC inscribed in a circle such as its hypotenuse forms the diameter AB of the circle (the right vertex is thus at the "apex" of the circle). If we procede to an infinitesimal displacement of the "apex" along the circumference $\ ds$ with subsequent horizontal and vertical displacements $\ dx$ and $\ dy$ , how does the function defined as $\ D(x)=a(x)+b(x)$ (with a and b the lengths of the two legs of the triangle) change with respect to the change in abscissa $\ x$? 
A crude depiction:

 [/INST] Let's assume your circle to be the unit circle. Then you have
\begin{align*}
 y &= \sqrt{1-x^2} \\
 y^2 &= 1-x^2 \\
 a(x) &= \sqrt{(1+x)^2+y^2} = \sqrt{1+2x+x^2+1-x^2} = \sqrt{2\left(1+x\right)} \\
 b(x) &= \sqrt{(1-x)^2+y^2} = \sqrt{1-2x+x^2+1-x^2} = \sqrt{2\left(1-x\right)} \\
 D(x) &= a(x)+b(x) = \sqrt{2\left(1+x\right)}+\sqrt{2\left(1-x\right)}
\end{align*}
I don't see a way to make this expression much easier, and neither does Wolfram Alpha. But you were writing about an infinitesimal change, so let's continue:
\begin{align*}
D'(x) &= \frac{\mathrm d\,D}{\mathrm dx} = \frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{2-2x^2}} \\
D'(0) &= 0 \\
D''(x) &= \frac{\mathrm d^2\,D}{\mathrm dx^2} = -\frac1{2\sqrt2}\left(
\left(1+x\right)^{-\frac32}+
\left(1-x\right)^{-\frac32}
\right) \\
D''(0) &= -\frac{1}{\sqrt2} \\
D'''(0) &= 0
\end{align*}
This corresponds to the beginning of a series expansion:
\begin{align*}
 D(x) &= D(0) + \frac{D''(0)}{2}x^2 + O\left(x^4\right) = 2\sqrt2 + \frac{1}{2\sqrt2}x^2 + O\left(x^4\right)
\end{align*}
If you want more terms, ask Wolfram Alpha. You can also obtain a series in terms of $\mathrm ds$ instead of $\mathrm dx$, simply by substituting $x=\sin s$. But since the first difference is in the $x^4$ (resp. $s^4$) term, the above approximation won't notice the difference.
</s>