Kartik's question at Yahoo! Answers regarding differentiation of a continued fraction

Staff member

MarkFL

Staff member
Hello Kartik,

We are given:

$$\displaystyle y=\cfrac{x}{a+\cfrac{x}{b+\cfrac{x}{a+\cfrac{x}{b+ \cdots}}}}$$

and asked to find $$\displaystyle \frac{dy}{dx}$$ in terms of $y$ only.

We may choose to write:

$$\displaystyle y=\cfrac{x}{a+\cfrac{x}{b+y}}$$

Multiplying through by $$\displaystyle \frac{1}{b+y}$$ we obtain:

$$\displaystyle \frac{y}{b+y}=\frac{x}{a(b+y)+x}$$

Inverting both sides, then subtracting through by 1, we have:

$$\displaystyle \frac{b}{y}=\frac{a(b+y)}{x}$$

Solving for $x$, we obtain:

$$\displaystyle x=ay+\frac{a}{b}y^2$$

Differentiating with respect to $y$, we find:

$$\displaystyle \frac{dx}{dy}=a+\frac{2a}{b}y=\frac{a(b+2y)}{b}$$

Hence:

$$\displaystyle \frac{dy}{dx}=\frac{b}{a(b+2y)}$$

chisigma

Well-known member
... inverting both sides, then subtracting through by 1, we have...

$$\displaystyle \frac{b}{y}=\frac{a(b+y)}{x}$$

Solving for $x$, we obtain...
If You want to obtain $\displaystyle \frac{d y}{d x}$ in 'standard form' [i.e. as function of the only x...] You can solve respect to y obtaining...

$\displaystyle y= - \frac{b}{2} \pm \sqrt{\frac{b^{2}}{4} + \frac{b}{a} x}\ (1)$

... and then differentiate (1). Note from (1) that y(x) is a multivalue function...

Kind regards

$\chi$ $\sigma$