MHB Kartik's Diff. of Cont. Fraction Q @ Yahoo Answers

AI Thread Summary
The discussion centers on a calculus problem involving the differentiation of a continuous fraction. The equation given is y = x/(a + x/(b + x/(a + b + ...))). The user seeks to find dy/dx in terms of y, leading to the transformation of the equation into a solvable form. After manipulating the equation, it is shown that dy/dx = b/(a(b + 2y)). Additionally, a method to express y as a function of x is provided, highlighting that y(x) is a multivalued function. The thread concludes with a focus on the differentiation process and the implications of the multivalued nature of y.
MarkFL
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Here is the question:

Differentiation : Calculus : Thanks :)?

http://www.flickr.com/photos/97838434@N06/9241146888/sizes/c/in/photostream/

Help needed with "Q.18) " on the above link. Thanks in advance :)

I have posted a link there to this topic so the OP can see my work.
 
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Hello Kartik,

We are given:

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

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

We may choose to write:

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

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

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

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

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

Solving for $x$, we obtain:

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

Differentiating with respect to $y$, we find:

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

Hence:

$$\frac{dy}{dx}=\frac{b}{a(b+2y)}$$
 
MarkFL said:
... inverting both sides, then subtracting through by 1, we have...

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

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

$\chi$ $\sigma$
 
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