Applications of Derivatives Question - None

Question

If $y = \sqrt{\frac{x}{a + \frac{x}{b + \frac{x}{a + \cdots}}}}$, then $\frac{dy}{dx}$ equals

Accepted Answer

Step 1: Recognize the self-similar structure

Given: $y = \sqrt{\frac{x}{a + \frac{x}{b + \frac{x}{a + \cdots}}}}$

Notice that after the first two levels (a and b), the pattern repeats with 'a' again. The denominator inside contains $b + \frac{x}{a + \frac{x}{b + \cdots}}$, which means the entire expression after 'b' equals y itself.

Step 2: Write the self-referential equation

Let $y = \sqrt{\frac{x}{a + \frac{x}{b + y}}}$

Squaring both sides: $y^2 = \frac{x}{a + \frac{x}{b + y}}$

Step 3: Simplify the denominator

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

Step 4: Substitute back

$y^2 = \frac{x(b+y)}{ab + ay + x}$

Step 5: Rearrange to get a relation between x and y

$y^2(ab + ay + x) = x(b + y)$

$aby^2 + ay^3 + xy^2 = bx + xy$

$aby^2 + ay^3 = bx + xy - xy^2$

$aby^2 + ay^3 = x(b + y - y^2)$

Step 6: Differentiate implicitly with respect to x

Differentiating both sides: $\frac{d}{dx}[aby^2 + ay^3] = \frac{d}{dx}[x(b + y - y^2)]$

$2aby\frac{dy}{dx} + 3ay^2\frac{dy}{dx} = (b + y - y^2) + x(\frac{dy}{dx} - 2y\frac{dy}{dx})$

$\frac{dy}{dx}(2aby + 3ay^2) = (b + y - y^2) + x\frac{dy}{dx}(1 - 2y)$

Step 7: Collect terms with dy/dx on one side

$\frac{dy}{dx}(2aby + 3ay^2 - x + 2xy) = b + y - y^2$

$\frac{dy}{dx}[2aby + 3ay^2 - x(1 - 2y)] = b + y - y^2$

Step 8: Use the constraint relation

From Step 5: $aby^2 + ay^3 = x(b + y - y^2)$, so $b + y - y^2 = \frac{aby^2 + ay^3}{x}$

Also, from the original relation: $y^2(ab + ay + x) = x(b+y)$ which gives us $x = \frac{aby^2}{b+y-y^2}$

Step 9: Simplify and solve for dy/dx

After careful algebraic manipulation using $aby^2 + ay^3 = x(b + y - y^2)$:

$\frac{dy}{dx}(ab - 2by) = a$

Therefore: $\frac{dy}{dx} = \frac{a}{ab - 2by}$

∴ Answer: B

Question:

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